32. Introduce the congruence notation: Use $\frac{b}{a}(\bmod n)$ to denote $b a^{-1}(\bmod n)$, where $a^{-1}$ is the modular inverse of $a$ modulo $n$ (note that this requires $(a, n)=1$).

Notice that, under the above fractional congruence notation, we only need to prove: $f_{p}(x)-f_{p}(y) \equiv 0\left(\bmod p^{3}\right)$. Since
$$\begin{aligned} f_{p}(x)-f_{p}(y) & =\sum_{k=1}^{p-1}\left(\frac{1}{(p x+k)^{2}}-\frac{1}{(p y+k)^{2}}\right) \\ & =\sum_{k=1}^{p-1} \frac{p^{2}\left(y^{2}-x^{2}\right)+p \cdot 2 k(y-x)}{(p x+k)^{2}(p y+k)^{2}} \end{aligned}$$

Therefore, we only need to prove: $\square$
$$\sum_{k=1}^{p-1} \frac{p\left(y^{2}-x^{2}\right)+2 k(y-x)}{(p x+k)^{2}(p y+k)^{2}} \equiv 0\left(\bmod p^{2}\right) .$$

And
$$\begin{aligned} (p x+k)^{2}(p y+k)^{2} & \equiv\left(2 p x k+k^{2}\right)\left(2 p y k+k^{2}\right) \\ & \equiv 2 p k^{3}(x+y)+k^{4}\left(\bmod p^{2}\right), \end{aligned}$$

Hence, to prove (10) holds, we only need to prove:
$$\begin{aligned} & \sum_{k=1}^{p-1} \frac{p\left(y^{2}-x^{2}\right)+2 k(y-x)}{2 p(x+y) k^{3}+k^{4}} \equiv 0\left(\bmod p^{2}\right) \\ \Leftrightarrow & \sum_{k=1}^{p-1}\left(\frac{2(y-x)}{k^{3}}-\frac{3 p\left(y^{2}-x^{2}\right)}{k^{4}+2 p(x+y) k^{3}}\right) \equiv 0\left(\bmod p^{2}\right) \end{aligned}$$

Furthermore, we only need to prove the following congruences hold simultaneously:
$$\left\{\begin{array}{l} \sum_{k=1}^{p-1} \frac{1}{k^{3}} \equiv 0\left(\bmod p^{2}\right) \\ \sum_{k=1}^{p-1} \frac{1}{k^{4}+2 p(x+y) k^{3}} \equiv 0(\bmod p) \end{array}\right.$$

For (11), since
$$\begin{aligned} 2 \sum_{k=1}^{p-1} \frac{1}{k^{3}} & =\sum_{k=1}^{p-1}\left(\frac{1}{k^{3}}+\frac{1}{(p-k)^{3}}\right) \\ & =\sum_{k=1}^{p-1} \frac{p^{3}-3 p^{2} k+3 p k^{2}}{k^{3}(p-k)^{3}} \equiv \sum_{k=1}^{\rho-1} \frac{3 p}{k(p-k)^{3}}\left(\bmod p^{2}\right) \end{aligned}$$

Hence, we only need to prove: $\sum_{k=1}^{p-1} \frac{1}{k(p-k)^{3}} \equiv 0(\bmod p)$. By
$$k(p-k)^{3} \equiv-k^{4}(\bmod p)$$

We only need to prove: $\sum_{k=1}^{p-1} \frac{1}{k^{4}} \equiv 0(\bmod p)$. In fact, (12) also reduces to proving this.
Below, we prove: When $p>5, p$ is a prime, $\sum_{k=1}^{p-1} \frac{1}{k^{4}} \equiv 0(\bmod p)$.
By Lagrange's theorem, we know that $x^{4} \equiv 1(\bmod p)$ has at most 4 solutions modulo $p$, so there exists $s \in\{1,2, \cdots, p-1\}$ such that $s^{4} \neq 1(\bmod p)$.

Notice that, $\left\{1, \frac{1}{2}, \cdots, \frac{1}{p-1}\right\}$ and $\left\{s, \frac{s}{2}, \cdots, \frac{s}{p-1}\right\}$ are both reduced residue systems modulo $p$, thus,
$$\sum_{k=1}^{p-1} \frac{1}{k^{4}} \equiv \sum_{k=1}^{p-1}\left(\frac{s}{k}\right)^{4}=s^{4} \sum_{k=1}^{p-1} \frac{1}{k^{4}}(\bmod p)$$

Thus, combining $p \nmid\left(s^{4}-1\right)$, we get $\sum_{k=1}^{p-1} \frac{1}{k^{4}} \equiv 0(\bmod p)$.
The proposition is proved.