Answer: either $f(x) = c$, $h(x) = \begin{cases} 0, & x \neq 0, \\ a, & x = 0, \end{cases}$ where $a$ and $c$ are arbitrary constants or $f(x) = x + b$, $h(x) = x$, where $b$ is an arbitrary constant.

Let $h(0) = a$. Set $x = 0$ in the initial equation
$$f(x^2 + y h(x)) = x h(x) + f(xy) \quad \text{for all } x, y \in \mathbb{R}. \quad (*)$$
We obtain $f(a y) = f(0) = c$ for all $y$.
If $a \neq 0$, then $a y$ admits all real values, so the function $f(x) = c$ is identically constant. So the equation has the form $c = x h(x) + c$ or $x h(x) = 0$, therefore $h(x) = 0$ for $x \neq 0$, and $h(0) = a$ admits any value. As it is easy to verify the obtained pair of the functions $(f(x), h(x))$ satisfies $(*)$.

Let $a = 0$, i.e. $h(0) = 0$. Note that if $h(x_0) \neq x_0$ for some $x_0$ ($x_0 \neq 0$), then there exists $y_0$ such that $x_0^2 + y_0 h(x_0) = x_0 y_0$ (indeed, it suffices to set $y_0 = x_0^2/(x_0 - h(x_0))$). Setting $x = x_0$, $y = y_0$, in the initial equation, we obtain $x_0 h(x_0) = 0$, i.e. $h(x_0) = 0$. Now setting $x = x_0$ in the initial equation, we have $f(x_0^2) = f(x_0 y)$ for all $y$. Since $x_0 \neq 0$, we see that $x_0 y$ admits any real values, so $f(x) = c$ is identically constant. But this case is already considered above.

It remains to suppose that $h(x) = x$ for all $x$. Then $(*)$ can be rewritten in the form
$$f(x^2 + y x) = x^2 + f(x y) \quad \text{for all } x, y \in \mathbb{R}. \quad (**)$$
Let $f(0) = b$. Setting $y = 0$ in $(**)$, we obtain $f(x^2) = x^2 + b$ for all $x$, i.e. $f(x) = x + b$ for all nonnegative $x$. If we set $y = -x$, then we obtain $f(-x^2) = -x^2 + b$, i.e. $f(x) = x + b$ for all nonpositive $x$.