If $k \neq 1$, then substituting $y = 0$ into the original equation gives
$$(f(0))^2 = (k - 1)f(x).$$
Therefore $f(x)$ is a constant function, and substituting back into the original equation we can solve $f(x) = 0$ or $f(x) = \frac{1}{k-1}$.

If $k = 1$, then substituting $x = 0$ into the original equation gives
$$f(y)^2 = f(y^2).$$
From the equation above we can know that for all $y > 0$ we have $f(y) > 0$ and $f(1) = 0$ or $1 > f(x)$. Substituting the above back into the original problem:
$$\begin{align*} & f(x) + (f(y))^2 = f(x + y^2), \forall x, y \in \mathbb{R} \\ \Rightarrow \quad & f(x) + f(y^2) = f(x + y^2), \forall x, y \in \mathbb{R} \\ \Rightarrow \quad & f(x) + f(y) = f(x + y), \forall x, y \in \mathbb{R}, y \ge 0 \end{align*}$$
This is exactly a standard Cauchy equation, therefore $f(x) = cx$. From $f(1) = 0$ or $1 > f(1) = 0$ or $f(x) = x$.