Annoying to do, I'm also a clown. Anyway

Let's just do it super fast: Define $P(x,y)$ to be the problem's assertion. $$P(x,0)\implies f(0)=f(x)^2(f(0)+1)-1\implies f(x)^2(f(0)+1) \mbox{ is a constant }$$ It's not hard to see that if $f(x)^2$ is a constant $\implies f\equiv \pm1$ (just take $x=\sqrt2$) So let's look at the case where $f(0)=-1$ and $f$ is not a constant: $P(0,x) \mbox{ gives } f \mbox{ is even }$ $P(1,x)\implies f(1)^2(f(x)+1)=0, \forall\ x \in \mathbb{R}\implies f(1)=0$ $P(x,1)\implies f(x^2-1)=f(x)^2-1 (\cdot)$ And now... The first somewhat good substitution here: $\forall z\in (-1,\infty)$ Let's take $x$ such that $x^2-1=z$: $$P(x,y)\implies f(yz)=(f(z)+1)f(y)+f(z)=(f(z)+1)(f(y)+1)-1$$ ...Huh? Of course, we know what to do from here: Let $g:=f+1\implies g$ is multiplicative! Translating our previous conditions to the new function we get that $$(\cdot)\implies g(x^2-1)=(g(x)-1)^2\ge0$$ As $f$ is even, $g$ is even$\implies g(x)\ge0\ \forall\ x\in\mathbb{R}$, so $h(x):=\sqrt{g(x)}\in \mathbb{R}_{>0}$ $$\implies h(x^2-1)^2=(h(x)^2-1)^2\implies h(x^2-1)=|h(x^2)-1| \mbox{ (h is multiplicative too)}$$ If we let $z,y\in\mathbb{R}_{>0}$, let $x^2y=z$: $$\implies h(y-z)=h(x^2-1)h(y)=h(y)|h(x^2)-1|=|h(y)-h(z)|$$ Let $Q(y,z)$ denote this assertion. Let $x,d\in \mathbb{R}_{>0}$, $$Q(x+2d,x+d)-Q(x+d,x)\implies |h(x+2d)-h(x+d)|=|h(x+d)-h(x)|$$ If $h(x+d)-h(x)=h(x+d)-h(x+2d)\implies h(2d)=|h(x+2d)-h(x)|=0$. As $h$ is multiplicative $\implies h\equiv 0$, a contradiction. So $h(x+d)-h(x)=h(x+2d)-h(x+d)$ for all $x,d\in \mathbb{R}_{>0}$. This is jensen's, and as $h(0)=0\implies h(x+y)=h(x)+h(y)$ As $h$ is both additive and multiplicative, we have $h\equiv x\implies f\equiv x^2-1$, which works edit: $h\equiv x$ for $x\ge0$, as we proved that it is cauchy only for $x,y\ge0$, but as $h$ is even, we have $h(-x)=h(x)\implies h\equiv |x|$