Okay, so I need to find the maximum value of the expression (x - y)/(x⁴ + y⁴ + 6) where x and y are real numbers. Hmm, let's start by understanding the problem. I need to maximize this fraction. Both the numerator and the denominator are functions of x and y, so I need to see how they relate.

First, maybe I can try to analyze the behavior of the function. Since it's a fraction, to maximize it, the numerator should be as large as possible relative to the denominator. But both the numerator and denominator can vary depending on x and y. Let me think about possible approaches.

One idea is to use calculus, taking partial derivatives with respect to x and y, setting them to zero, and solving for critical points. But that might get complicated because the denominator is a fourth-degree polynomial, and the numerator is linear. Let's see.

Alternatively, maybe I can use substitution. Let me set u = x - y and v = x + y. Then, maybe express the denominator in terms of u and v. But I'm not sure if that helps immediately. Let's compute:

x = (u + v)/2 and y = (v - u)/2. Then, x⁴ + y⁴ would be [(u + v)/2]^4 + [(v - u)/2]^4. That seems messy. Maybe another substitution?

Wait, perhaps symmetry can help here. The denominator is symmetric in x and y because swapping x and y would result in the same expression. But the numerator is anti-symmetric; swapping x and y changes its sign. So maybe the maximum occurs when x and y are related in a specific way.

Alternatively, maybe set t = x - y, and then express the denominator in terms of t and some other variable. Let's try that. Let t = x - y, so y = x - t. Then substitute into the denominator:

x⁴ + (x - t)⁴ + 6. Hmm, but then we still have two variables, x and t. Maybe not helpful.

Wait, perhaps consider fixing t and then optimizing over x. Or maybe use some inequality to bound the expression.

Let me think about inequalities. The denominator has x⁴ + y⁴ + 6. Since x⁴ and y⁴ are always non-negative, the denominator is at least 6. The numerator is x - y, which can be positive or negative. So the maximum value would be when the numerator is positive and as large as possible relative to the denominator.

Alternatively, maybe use the AM-GM inequality. But the denominator is a sum of fourth powers, which complicates things. Wait, but perhaps I can bound x⁴ + y⁴ from below using some relation to (x - y)² or something like that.

But (x - y)² = x² - 2xy + y², which is different from x⁴ + y⁴. Not sure. Maybe Cauchy-Schwarz?

Alternatively, maybe use the method of Lagrange multipliers. Let me try that. Let me set f(x, y) = (x - y)/(x⁴ + y⁴ + 6). To maximize f, set up the gradient equal to zero.

So, compute the partial derivatives of f with respect to x and y. Let's compute ∂f/∂x:

Using quotient rule: [1*(x⁴ + y⁴ + 6) - (x - y)*(4x³)] / (x⁴ + y⁴ + 6)^2.

Similarly, ∂f/∂y: [ -1*(x⁴ + y⁴ + 6) - (x - y)*(4y³) ] / (x⁴ + y⁴ + 6)^2.

Set both partial derivatives equal to zero. Therefore, the numerators must be zero.

So:

1*(x⁴ + y⁴ + 6) - (x - y)*(4x³) = 0,

and

-1*(x⁴ + y⁴ + 6) - (x - y)*(4y³) = 0.

Let me denote the first equation as Equation (1):

x⁴ + y⁴ + 6 - 4x³(x - y) = 0,

and the second equation as Equation (2):

-x⁴ - y⁴ -6 -4y³(x - y) = 0.

If I add Equations (1) and (2):

(x⁴ + y⁴ +6 -4x³(x - y)) + (-x⁴ - y⁴ -6 -4y³(x - y)) = 0.

Simplify:

0 + 0 + (-4x³(x - y) -4y³(x - y)) = 0.

Factor out -4(x - y):

-4(x - y)(x³ + y³) = 0.

So, either x - y = 0 or x³ + y³ = 0.

Case 1: x - y = 0. Then x = y. Substitute into the original function:

(x - x)/(x⁴ + x⁴ +6) = 0/(2x⁴ +6) = 0. So the function is zero here. So not the maximum.

Case 2: x³ + y³ = 0. So x³ = -y³. Therefore, y = -x. Since x and y are real numbers, this implies y = -x. So substitute y = -x into the original function.

Then, numerator becomes x - (-x) = 2x.

Denominator becomes x⁴ + (-x)⁴ +6 = 2x⁴ +6.

So the function is 2x / (2x⁴ +6) = x / (x⁴ +3).

So now, we need to maximize f(x) = x/(x⁴ +3) for real x. Let's compute derivative of f(x):

f'(x) = [1*(x⁴ +3) - x*(4x³)] / (x⁴ +3)^2 = (x⁴ +3 -4x⁴)/(x⁴ +3)^2 = (-3x⁴ +3)/(x⁴ +3)^2.

Set derivative equal to zero:

-3x⁴ +3 = 0 => x⁴ =1 => x = ±1.

So critical points at x=1 and x=-1.

Compute f(1) = 1/(1 +3) = 1/4.

f(-1) = -1/(1 +3) = -1/4.

So maximum value is 1/4 at x=1, y=-1.

But wait, we need to check if this is indeed the global maximum. Also, check the behavior as x approaches infinity.

As x approaches infinity, f(x) = x/(x⁴ +3) ~ 1/x³ approaches zero. Similarly, as x approaches negative infinity, f(x) approaches zero from the negative side. So the maximum is indeed at x=1, y=-1 with value 1/4.

But wait, let's verify that this is the global maximum. Since the function f(x) = x/(x⁴ +3) has only two critical points at x=1 and x=-1, and the endpoints go to zero, these are indeed the global max and min.

Therefore, the maximum value of the original expression is 1/4. But let's check if there are any other critical points in the original function when using Lagrange multipliers.

Wait, in the case where x³ + y³ =0, we found y = -x, leading to the function x/(x⁴ +3). But in the Lagrange multiplier method, we considered both cases. So that's all the critical points.

Therefore, the maximum value is 1/4. Let me check with some test values.

For example, if x=1, y=-1: numerator is 1 - (-1)=2, denominator is 1 +1 +6=8, so 2/8=1/4. Correct.

If x=2, y=-2: numerator 4, denominator 16 +16 +6=38, so 4/38≈0.105, which is less than 1/4.

If x=0, y=0: numerator 0, denominator 6, value 0.

If x=1/2, y=-1/2: numerator 1, denominator (1/16) + (1/16) +6=1/8 +6=49/8, so 1/(49/8)=8/49≈0.163, which is less than 1/4.

So seems 1/4 is indeed the maximum.

Wait, but let me check another critical point. Suppose we didn't use Lagrange multipliers, maybe there's another critical point.

But according to the Lagrange multiplier method, the only critical points are when x=y (which gives zero) or when x³ + y³=0, leading to y=-x. So that's all. Therefore, the maximum is 1/4.

Hence, the answer should be 1/4.

**Final Answer**
The maximum value is \boxed{\dfrac{1}{4}}.
Therefore, the maximum value of the given expression is $\boxed{\dfrac{1}{4}}$.
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