Determine all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for every $x,y \in \mathbb{R}$ 
$$f(x^{4}-y^{4})+4f(xy)^{2}=f(x^{4}+y^{4})$$
1. Let \( P(x, y) \) be the assertion \( f(x^4 - y^4) + 4f(xy)^2 = f(x^4 + y^4) \).

2. Consider \( P(0, 0) \):
   \[
   f(0^4 - 0^4) + 4f(0 \cdot 0)^2 = f(0^4 + 0^4) \implies f(0) + 4f(0)^2 = f(0) \implies 4f(0)^2 = 0 \implies f(0) = 0.
   \]

3. Consider \( P(0, x) \):
   \[
   f(0^4 - x^4) + 4f(0 \cdot x)^2 = f(0^4 + x^4) \implies f(-x^4) + 4f(0)^2 = f(x^4) \implies f(-x^4) = f(x^4).
   \]
   This shows that \( f \) is an even function.

4. Consider \( P(x, x) \):
   \[
   f(x^4 - x^4) + 4f(x \cdot x)^2 = f(x^4 + x^4) \implies f(0) + 4f(x^2)^2 = f(2x^4) \implies 4f(x^2)^2 = f(2x^4).
   \]
   Since \( f(0) = 0 \), we have \( f(2x^4) = 4f(x^2)^2 \). This implies \( f(x) \geq 0 \) for all \( x \).

5. Define \( h(x) = \sqrt{f(x)} \). Since \( f \) is nonnegative and even, \( h \) is also nonnegative and even.

6. From \( f(2x^4) = 4f(x^2)^2 \), we get:
   \[
   h(2x^4) = 2h(x^2).
   \]

7. Since \( f(x) = \frac{1}{2}h(2x^2) \) for all \( x \geq 0 \), and this holds for \( x < 0 \) as well due to the even nature of \( f \).

8. Rewrite \( P(x, y) \) using \( h \):
   \[
   f(x^4 - y^4) + f(2x^2y^2) = f(x^4 + y^4).
   \]
   Using \( f(2x^4) = 4f(x^2)^2 \), we get:
   \[
   h(2(x^4 - y^4)^2) + h(8x^4y^4) = h(2(x^4 + y^4)^2).
   \]
   Simplifying, we have:
   \[
   h((x - y)^2) + h(4xy) = h((x + y)^2).
   \]

9. This implies \( h(u) + h(v) = h(u + v) \) for all \( u, v \geq 0 \). Since \( h \) is additive and nonnegative, it must be of the form \( h(x) = cx \) for some constant \( c \geq 0 \).

10. Therefore, \( f(x) = \frac{1}{2}h(2x^2) = cx^2 \) for all \( x \).

11. Plugging this back into the original equation, we get:
    \[
    c(x^4 - y^4) + 4(cxy)^2 = c(x^4 + y^4).
    \]
    Simplifying, we find:
    \[
    c(x^4 - y^4) + 4c^2x^2y^2 = c(x^4 + y^4).
    \]
    This implies:
    \[
    c(x^4 - y^4) + 4c^2x^2y^2 = c(x^4 + y^4) \implies 4c^2x^2y^2 = 2cy^4 \implies 2c^2x^2y^2 = cy^4.
    \]
    For this to hold for all \( x \) and \( y \), we must have \( c = 0 \) or \( c = 1 \).

12. Therefore, the solutions are:
    \[
    \boxed{f(x) = 0 \quad \forall x} \quad \text{and} \quad \boxed{f(x) = x^2 \quad \forall x}.
    \]