Example 9 Given that $f(x)$ is a function defined on the set of natural numbers $\mathrm{N}$, satisfying $f(1)=\frac{3}{2}$, and for any $x, y \in \mathrm{N}$, 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 $f(x)$.

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Solution: In the original functional equation, let $y=1$, and use $f(1)=\frac{3}{2}$, we get $f(x+1)=\left(1+\frac{1}{x+1}\right) f(x)+\left(1+\frac{x}{2}\right) \cdot \frac{3}{2}+x^{2}+2 x$.
Rearranging gives $\frac{f(x+1)}{x+2}-\frac{f(x)}{x+1}=x+\frac{3}{4}$.
Let $x=1,2, \cdots, n-1$ to get
$$
\begin{array}{l}
\frac{f(2)}{3}-\frac{f(1)}{2}=1+\frac{3}{4} \\
\frac{f(3)}{4}-\frac{f(2)}{3}=2+\frac{3}{4} \\
\cdots \cdots
\end{array}
$$
$$
\frac{f(n)}{n+1}-\frac{f(n-1)}{n}=(n-1)+\frac{3}{4}
$$

Adding the above equations, we get
$$
\frac{f(n)}{n+1}-\frac{f(1)}{2}=\frac{1}{2}(n-1) n+\frac{3}{4}(n-1)=\frac{1}{4}(n-1)(2 n+3) \text {. }
$$

Substituting $f(1)=\frac{3}{2}$ and rearranging, we get
$$
f(n)=\frac{1}{4} n(n+1)(2 n+1) \text {. }
$$

Thus, the required function is $f(x)=\frac{1}{4} x(x+1)(2 x+1)$ for $x \in \mathbb{N}$.
It is easy to verify that $f(x)$ satisfies the original functional equation.