let’s break down the functional equation you’ve got:

\[ f(x^4 - y^4) + 4f(xy)^2 = f(x^4 + y^4). \]

First, notice that \( x^4 - y^4 \) can be factored into \( (x^2 - y^2)(x^2 + y^2) \), and \( x^4 + y^4 \) can also be expressed in terms of squares. This suggests that we might want to explore specific values for \( x \) and \( y \) to find patterns or potential solutions.

A good starting point is to plug in some simple values. For instance, let’s set \( y = 0 \):

\[ f(x^4) + 4f(0)^2 = f(x^4). \]

This simplifies to \( 4f(0)^2 = 0 \), which means \( f(0) = 0 \).

Next, if we let \( x = y \):

\[ f(0) + 4f(x^2)^2 = f(2x^4). \]

Since we already found that \( f(0) = 0\), this reduces to:

\[ 4f(x^2)^2 = f(2x^4). \]

Now, if we assume a potential solution form like \( f(x) = cx \), where \( c \) is a constant, we can substitute it back into the original equation and check if it holds. However, you'll find that a linear function doesn’t satisfy the equation generally unless \( c = 0\).

After some trials and analyzing the structure, a plausible solution emerges: **the zero function** \( f(x) = 0 \) for all \( x \in \mathbb{R} \). This indeed satisfies the original equation.

In summary, after examining various substitutions and potential forms for \( f \), the conclusion is that the only function that fits is:

\[ f(x) = 0. \]

S. Stewart