See the direct formula for the probability density function (pdf) here:
https://en.wikipedia.org/wiki/Distribution_of_the_product_of_two_random_variables
Here's the standard proof that only uses the change-of-variables formula from multivariate calculus. It's just a flattening of the arguments of the other answers above to something elementary.
Let $X$ and $Y$ be independent random variables with $\mathbb{P}(Y=0) = 0$.
Write $T = X \cdot Y$ and $U = Y$. Observe $g(T,U) = (X,Y)$ where $g(t,u) := (t/u, u)$.
Then, in terms of the jacobian matrix $\partial g = \begin{bmatrix} 1/u & -t/u^2\\ 0 & 1\end{bmatrix}$, note the joint pdf $$ f_{T,U}(t,u) = f_{X,Y}(g(t,u)) \cdot | \mathrm{det}\,\partial g | = f_X(t/u) \cdot f_Y(u) \,/\, |u|.$$
Therefore, we obtain the desired pdf as its marginal pdf via ``partial integration'': $$ f_{X\cdot Y}(t) = \int_{\mathbb{R}} f_{T,U}(t,u) \, \partial u = \int_{-\infty}^\infty f_X\left(\frac{t}{u}\right) \cdot f_Y(u) \, \frac{\partial u}{u}. $$
This method is generic and applies to finding the pdf of $\varphi(X,Y)$ for any $C^1$-function $\varphi:\mathbb{R}^2 \longrightarrow \mathbb{R}$.