Example 3 If the function $f(x)$ satisfies
$$
\begin{array}{l}
f(x+y)=f(x)+f(y)+x y(x+y), \\
f^{\prime}(0)=1,
\end{array}
$$

find the analytical expression of the function $f(x)$.
Notice that, $-x-y+(x+y)=0$.
From formula (1) we get
$$
\begin{array}{l}
(-x)^{3}+(-y)^{3}+(x+y)^{3} \\
=3(-x)(-y)(x+y)=3 x y(x+y) . \\
\text { By } f(x+y)=f(x)+f(y)+x y(x+y) \\
\Rightarrow f(x+y)=f(x)+f(y)+\frac{1}{3}\left[(x+y)^{3}-x^{3}-y^{3}\right] \\
\Rightarrow f(x+y)-\frac{1}{3}(x+y)^{3} \\
\quad=f(x)-\frac{1}{3} x^{3}+f(y)-\frac{1}{3} y^{3} .
\end{array}
$$

Let $g(x)=f(x)-\frac{1}{3} x^{3}$. Then the above equation becomes
$$
g(x+y)=g(x)+g(y) \text {. }
$$

Since $f^{\prime}(0)=1$, $f(x)$ is continuous at $x=0$.
Thus, equation (3) is a Cauchy equation, whose solution is $g(x)=a x(a=g(1))$.
Therefore, $f(x)=\frac{1}{3} x^{3}+a x \Rightarrow f^{\prime}(x)=x^{2}+a$.
By $f^{\prime}(0)=1$, we know $a=1$.
Hence, $f(x)=\frac{1}{3} x^{3}+x$.