= Solution
For a <Borel set> $A\subseteq\mathbb R$, the <joint probability density> gives
$$
\mathbb P(XY\in A)=\int_{\mathbb R}\int_{\mathbb R}\mathbf1_A(xy)f(x,y)\,dy\,dx.
$$
For each $x\ne0$, substitute $z=xy$. The <change of variables for probability densities> contributes $dy=dz/|x|$, with the absolute value accounting for the reversed orientation when $x<0$. The line $x=0$ has two-dimensional <Lebesgue measure> zero, so omitting it changes no <probability>. By the <Tonelli theorem>,
$$
\mathbb P(XY\in A)=\int_A\left[\int_{\mathbb R\setminus\{0\}}\frac1{|x|}f\left(x,\frac zx\right)\,dx\right]dz.
$$
The bracket is nonnegative and its <integral> over $\mathbb R$ is one. This proves the <probability density of a product of random variables> formula:
$$
\boxed{f_Z(z)=\int_{\mathbb R\setminus\{0\}}\frac1{|x|}f\left(x,\frac zx\right)\,dx.}
$$
As usual, a <probability density function> is determined only almost everywhere; exceptional infinite values are harmless.
For the <independent random variables> with the specified <uniform distribution>, $f(x,y)=b^{-2}$ on $(0,b)^2$ and zero elsewhere. When $0<z<b^2$, the conditions $0<x<b$ and $0<z/x<b$ are equivalent to $z/b<x<b$. Therefore
$$
f_Z(z)=\frac1{b^2}\int_{z/b}^{b}\frac{dx}{x}=\frac1{b^2}\log\frac{b^2}{z},
\qquad
\boxed{f_Z(z)=\begin{cases}b^{-2}\log(b^2/z),&0<z<b^2,\\0,&\text{otherwise}.\end{cases}}
$$
The value assigned at zero is arbitrary. Substitution $s=z/b^2$ checks <normalization of a probability distribution>: $\int_0^1-\log s\,ds=1$. This is the <product of independent uniform random variables> law.
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