jmc

Setting $x=y=0,$ we get $$f(0{)}^2=4f(0{)}^2.$$Then $f(0{)}^2=0,$ so $f(0)=0.$

Setting $x=y,$ we get $$4f(x{)}^2-4x^{2}f(x)=0,$$so $f(x)(f(x)-x^2)=0.$ This tells us that for each value of $x,$ either $f(x)=0$ or $f(x)=x^2.$ (Note that it does not tell us that either $f(x)=0$ for all $x,$ or $f(x)=x^2$ for all $x.$)

We can easily check that $f(x)=x^2$ satisfies the given functional equation. Otherwise, there exists some nonzero real number $a$ such that $f(a)=0.$ Setting $y=a,$ we get $$f(x+a)f(x-a)=f(x{)}^2$$for all $x.$ Suppose there exists a real number $b$ such that $f(b)\ne 0.$ Then $f(b)=b^2.$ Substituting $x=b$ into the equation above, we get $$f(b+a)f(b-a)=f(b{)}^2=b^4.$$Since $f(b)=b^2\ne 0,$ both $f(b+a)$ and $f(b-a)$ must be nonzero. Therefore, $f(b+a)=(b+a{)}^2$ and $f(b-a)=(b-a{)}^2,$ and $$(b+a{)}^2(b-a{)}^2=b^4.$$Expanding, we get $a^4-2a^2{b}^2+b^4=b^4,$ so $a^4-2a^2{b}^2=0$. Then $a^2(a^2-2b^2)=0.$ Since $a$ is nonzero, $a^2=2b^2,$ which leads to $b=\pm \frac{a}{\sqrt{2}}.$

This tells us that if there exists some nonzero real number $a$ such that $f(a)=0,$ then the only possible values of $x$ such that $f(x)\ne 0$ are $x=\pm \frac{a}{\sqrt{2}}.$ We must have that $f(x)=0$ for all other values of $x.$ We can then choose a different value of $a^{\prime }$ such that $f(a^{\prime })=0,$ which leads to $f(x)=0$ for all $x$ other than $x=\pm \frac{a^{\prime }}{\sqrt{2}}.$ This forces $f(x)=0$ for all $x,$ which easily satisfies the given functional equation.

Therefore, there are only $\boxed{2}$ functions that work, namely $f(x)=0$ and $f(x)=x^2.$