Find all functions $f:\mathbb R\to\mathbb R$ such that, for all $x,y\in\mathbb R$, we have $$f(x^2-y^2)=f(xy)(f(x)+f(y)).$$
Both $f(x)=0$ and $f(x)=1/2$ are solutions.
Swapping $x,y$ in the equation implies that $f$ is even. Plugging in $x=y=0$ implies that $f(0)=0$ or $f(0)=1/2$.
If $f(0)=0$, then plugging in $y=0$ implies that $f(x^2)=0$. So $f(t)=0$ for $t\geq0$, and $f$ even means that $f=0$ is the only solution in this case.
I'm struggling though with the case $f(0)=1/2$.
Plugging in $y=0$ gives $f(x^2)=f(0)(f(x)+f(0))$, which, after plugging in $f(0)=\frac 1 2$, gives $$f(x^2)=\frac 1 2 f(x) + \frac 1 4$$ on the other hand, plugging in $x=y$ gives $f(0)=f(x^2)(f(x)+f(x))$, which is $$\frac 1 2=2f(x)f(x^2)$$ Eliminating $f(x^2)$ in those two equations provides $$2(f(x))^2+f(x)-1=0$$ which implies either $f(x)=\frac 1 2$ or $f(x)=-1$. Since $f(0)=\frac 1 2$, we can safely conclude that $f=\frac 1 2$ is the solution.