Put $x=-1$ to see that $f(-1)^2+2h(-1)^2=0$. Hence $f(-1)=h(-1)=0$ since both are real. But then the RHS is divisible by $(x+1)^2$ as well as $f(x)^2$, hence also $(x+1)g(x)^2$ is divisible by $(x+1)^2$ and so $g(x)$ is divisible by $x+1$ as well. But then dividing everything by $(x+1)^2$ we get another solution. Starting with the solution of minimal positive degree, we therefore get a contradiction. Hence the only solution is $f(x)=g(x)=h(x)=0$.