Example 6 If the function $f(x)$ satisfies
$$
\begin{array}{l}
f(x+y)=f(x)+f(y)+xy(x+y), \\
f^{\prime}(0)=1,
\end{array}
$$

find the analytical expression of the function $f(x)$.
(2000, Shanghai Jiao Tong University Admissions Exam for Recommended Students)
Notice,
$$
\begin{array}{l}
x y(x+y)=(-x)(-y)(x+y), \\
-x-y+(x+y)=0 \\
\text { Hence }(-x)^{3}+(-y)^{3}+(x+y)^{3}=3 x y(x+y) . \\
\text { By } f(x+y)=f(x)+f(y)+x y(x+y) \\
\Rightarrow f(x+y) \\
\quad=f(x)+f(y)+\frac{1}{3}\left[(x+y)^{3}-x^{3}-y^{3}\right] \\
\Rightarrow f(x+y)-\frac{1}{3}(x+y)^{3} \\
\quad=f(x)-\frac{1}{3} x^{3}+f(y)-\frac{1}{3} y^{3} .
\end{array}
$$

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

By $f^{\prime}(0)=1$, then $f(x)$ is continuous at $x=0$.
Thus, equation (2) is a Cauchy functional 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 $a=1$.
Thus, $f(x)=\frac{1}{3} x^{3}+x$.