Given the answer I got, I computed the PDF of $Z=XY$, where $X,Y\sim U(0,1)$ and independent. Using the formula:
$$f_Z(z)=\int_{-\infty}^{\infty}f_X(x)f_Y(z/x){dx\over |x|}$$
$$=-\int_{-\infty}^0f_X(x)f_Y(z/x){dx\over x}+\int_0^{\infty}f_X(x)f_Y(z/x){dx\over x}$$
When $z\leq 0$, it would reflect $f_Y(z/x)$, so $f_X(x)f_Y(z/x)=0$ and for $x\leq 0$
$$-\int_{-\infty}^0f_X(x)f_Y(z/x){dx\over x}=0$$
The scaling of the argument in $f_Y(z/x)$ transform the domain where is $1$. For $z\geq 0$, I analize $f_Y(y)$ as $H(y)-H(y-1)$, where $H(y)$ is the Heaviside step function. Manipulating the inequalities for $H(z/x)$:
$${z\over x}<0\Rightarrow {x\over z} < 0\Rightarrow x<0 $$
$${z\over x} \geq 0\Rightarrow {x\over z}\geq 0\Rightarrow x\geq 0 $$
indicates that $H(z/x)=H(x)$. Now manipulating the inequalities for $H(z/x -1)$:
$${z\over x}-1<0\Rightarrow {x\over z} > 1\Rightarrow x>z $$
$${z\over x}-1 \geq 0\Rightarrow {x\over z}\leq 1\Rightarrow x\leq z $$
From this, I'd argue that $H(z/x-1)=H(-x+z)$ but graphing it in Desmos indicates that $H(z/x-1)=H(x)-H(x-z)$. In this case, evaluating when $x\geq 0$ gives the same result. Now, by definition of $f_Y(z/x)$
$$f_Y(z/x)=H(z/x)-H(z/x-1)=H(x)-(H(x)-H(x-z))=H(x-z)$$
$f_X(x)f_Y(z/x)$ isn't $0$ when $z\leq x\leq 1$, so the integral to evaluate is:
$$\int_0^{\infty}f_X(x)f_Y(z/x){dx\over x}=\int_z^1 {1\over x}dx=-\ln(z)=f_Z(z)$$
$-\ln(z)$ is positive for $z\in[0,1]$ and $\int_{[0,1]}-\ln(z)dz=1$ so i think I got it right. Is this true?
Since you know the convolution, a nice alternative to this is to consider the exponentially distributed variables $\log X$ and $\log Y$ whose sum is $\log Z=\log X + \log Y$. If $X,Y\sim U(0,1)$, then, for $a\in (-\infty,0]$,
$$P(\log X<a)=P(X<e^a)=e^a$$
so the pdf of $\log X$ is $e^x$ when $x<0$ and $0$ for $x>0$. Of course the pdf of $\log Y$ is the same. Convolving these gives the pdf $f_{\log Z}$ of $\log Z$. Finally, for $a\in [0,1]$,
$$P(Z<a)=P(\log Z<\log a)=\int_{-\infty}^{\log a} f_{\log Z}(t)dt=\int_0^{a} f_{\log Z}(\log t)\frac{dt}{t}$$ so the pdf of $Z$ is $\frac{f_{\log Z}(\log t)}{t}$.