Problems from Ukrainian Authors

Let us denote the equation given in the statement of a problem by $(*)$ and substitute $y = -x$ into it. We obtain
$$f^2(x) + f^2(-x) + 2f(-x^2) = f^2(0) \text{\{ (**).\}}$$
Substituting $x = x + y, y = -x$ into $(*)$, we get
$$f^2(y) = f^2(x + y) + f^2(-x) + 2f(-x^2 - xy).$$
Adding $(*)$ and $(**)$ to the last-mentioned equality, we obtain
$$f(xy) + f(-x^2 - xy) = f(-x^2) - f^2(0)/2.$$
Let us denote $g(x) = f(x) + f^2(0)/2$. Let $a = -x^2 \le 0, b = xy \in \mathbb{R}$ when $a \ne 0$. The last-mentioned equality assumes the form
$$g(a) + g(b) = g(a + b) \text{\{ (***)\}.}$$
If $a, b < 0$, then we obtain the Cauchy's equation. Let us prove that $g(x)$ is bounded above when $x < 0$. Then from the Cauchy's equation we get that $g(x) = kx$ for arbitrary negative number $x$ and real number $k$. From $(**)$ we get

Thus, $g(x) = kx \quad \forall x < 0$. Let us assume that $x \ge 0$. Substituting $y = -1 - x$ into $(***)$, we obtain $g(x) = g(-1) - g(-1 - x) = kx$ for every positive number $x$. Thus, we proved that
$$f(x) = g(x) - f^{2}(0)/2 = kx - f^{2}(0)/2 = kx + c$$
for every real $x$. Substituting the explicit form of the function $f$ into $(*)$, we obtain:
$$\begin{align*} k^2 (x + y)^2 + 2kc(x + y) + c^2 &= k^2 x^2 + 2kcx + c^2 + 2kxy + 2c + k^2 y^2 + 2kcy + c^2, \\ (k^2 - k)xy &= c^2 + 2c. \end{align*}$$
Considering that $xy$ is able to possess arbitrary real value, from the last-mentioned equality we get that $k \in \{0,1\}$, $c \in \{0,-2\}$.
Thus, we obtain that the condition can be fulfilled only by the functions $f(x) = 0$, $f(x) = -2$, $f(x) = x$, and $f(x) = x - 2$. We verify that all these functions fulfill the condition of the problem by checking.