Four. (50 points) Given that $f(x)$ is a function defined on the set of natural numbers $\mathbf{N}^{\boldsymbol{}}$, satisfying $f(1)=\frac{3}{2}$, and for any $x$, $y \in \mathbf{N}^{*}$, there is $f(x+y)=\left(1+\frac{y}{x+1}\right) f(x)+\left(1+\frac{x}{y+1}\right) f(y)+x^{2} y+x y+x y^{2}$, find $f(x)$.


Four, Solution: Substituting $y=1$ yields $f(x+1)=\left(1+\frac{1}{x+1}\right) f(x)+\left(1+\frac{x}{2}\right) f(1)+x^{2}+2 x$,
rearranging gives $\frac{f(x+1)}{x+2}-\frac{f(x)}{x+1}=x+\frac{3}{4}$. Let $x=1,2, \cdots, n-1$, we get $\frac{f(2)}{3}-\frac{f(1)}{2}=1+\frac{3}{4}, \frac{f(3)}{4}-\frac{f(2)}{3}=2+\frac{3}{4}, \cdots, \frac{f(n)}{n+1}-\frac{f(n-1)}{n}=(n-1)+\frac{3}{4}$. Adding all these equations, we get $\frac{f(n)}{n+1}-\frac{f(1)}{2}=\frac{1}{2} n(n-1)+\frac{3}{4}(n-1)$.
Rearranging gives $f(n)=\frac{1}{4} n(n+1)(2 n+1)$, the required function is $f(x)=\frac{1}{4} x(x+1)(2 x+1), x \in \mathbf{N}^{*}$.