Q:

How to calculate this joint distribution? 

Suppose $f:\mathbb{R}\to[0,\infty)$ is Borel measurable and the integral of $f$ is strictly positive and finite.  Moreover we have a r.v. 's $X$ and  $Y$, where $Y$ is uniformly(0,1) distributed and independent of $X$. Furthermore we assume that the distribution of $X$ is given by
$$P_X(A)=\frac{\int_Af(x)dx}{\int_\mathbb{R}f(x)dx}$$
Then I want to calculate the distribution Function of $Z:=(X,f(X)Y)$, i.e.
$$P(Z\in A)$$
for a Borel set $A=A_1\times A_2\in \mathbb{R}\times\mathbb{R}$.
$$P(Z\in A)=P(X\in A_1,f(X)Y\in A_2)$$
I know that $X$ and $Y$ are independent, but I think $X,f(X)Y$ are not in general.  So how can I compute this?  It's an exercise in my probability book.  The solution should be, that $Z$ is uniformly subgraph$(f)$ distributed, which means. 
$$\mathcal{U}_{subgraph(f)}(B)=\frac{\lambda(B)}{\lambda(subgraph(f))}$$
where $\lambda$ is the lebesgue measure and $subgraph(f):=\{(x,y)\in \mathbb{R}^2;0\le y\le f(x)\}$. I know that $\lambda(subgraph(f))=\int_\mathbb{R}f(x)dx$.

A:

For brevity, we will assume that $\int_{\mathbb R} f\, d\lambda=1$ (just replace $f$ by $f/\int_{\mathbb R} f\, d\lambda$ otherwise). 
Given $X = x$, we now, that $f(X)Y$ is uniformly $[0, f(x)]$-distributed.  Hence the density of $Z$ is given by 
\begin{align*}
  d_Z(z_1, z_2) &= d_{f(X)Y}(z_2 \mid X = z_1)d_X(z_1)\\
                &= \frac 1{f(z_1)}\chi_{[0,f(z_1)]}(z_2) \cdot f(z_1)\\
                &= \chi_{[0,f(z_1)]}(z_2)
\end{align*}
Now, for a Borel set $A \subseteq \mathbb R^2$, we have 
\begin{align*}
  P(Z \in A) &= \int_A d_Z(z)\, dz\\
             &= \int_A \chi_{[0,f(z_1)]}(z_2)\, dz\\
             &= \int_{\mathbb R^2} \chi_A(z)\chi_{[0,f(z_1)]}(z_2)\, dz\\
             &= \int_{\mathbb R^2} \chi_{\{z \in A \mid 0 \le z_2 \le f(z_1)\}}(z)\, dz\\
             &= \lambda^2(\{z \in A \mid 0 \le z_2 \le f(z_1)\}). 
\end{align*}
which is the portion of the subgraph of $f$ contained in $A$ (your denominator above vanishes as I supposed $\int_{\mathbb R} f\, d\lambda^1 = \lambda^2(\text{subgraph}\, f) = 1$.

