$X_1$ and $X_2$ is exponentially distributed with the parameter 1. I need to find the density function of $(X_1+X_2,X_1)$. The density function of $(X_1,X_2)$ is given by $p(x_1,x_2)=\exp(-(x+y))$.
Is it then correct that $p(x_1+x_2,x_1)=\exp(-(x_1+x_2))$
By a change of variables, $$\begin{align}p_{X_1+X_2, X_1}(s,t) ~=~& p_{X_2,X_1}(s-t,t) \\[1ex]~=~& p_{X_2}(s-t)\cdotp p_{X_1}(t) \\[1ex]~=~& \mathsf e^{-(s-t)}\cdotp \mathsf e^{-t}\cdotp \mathbf 1_{s\in(0;\infty)\,\land\,s-t\in (0;\infty)}\\[1ex]~=~&\mathsf e^{-s}\cdotp\mathbf 1_{0\leq t\leq s}\end{align}$$
This should really be a new question, but anyway: no. Among other issues, you should anticipate that $q(z)$ will be some function of $z$, which you made no attempt to include in your integration. Thus that is clearly not the correct answer.
Let's use $X,Y$ instead of $X_1,X_2$ for reasons of clarity and compactness.
If $Z=X/(X+Y)$ then $Y=(X/Z)-X$, (and the map $Y\mapsto Z$ is bijective for any given $X$ in their support) so the Jacobian change of variables transformation is:$$\begin{align}p_{Z, X}(z,x) ~=~& p_{Y,X}((x/z)-x,x)\cdotp\left\lvert\frac{\partial((x/z)-x,x)}{\partial(z,x)}\right\rvert \\[1ex] =~&p_Y((x/z)-x)\cdotp p_X(x)\cdotp\begin{Vmatrix}-x/z^2& (1/z)-1\\0 & 1 \end{Vmatrix}\\[1ex]~=~& \mathsf e^{x-x/z}\cdotp\mathsf e^{-x}\cdotp\Bigl\lvert\frac x{z^2}\Bigr\rvert\cdotp\mathbf 1_{(x/z)-x\in(0;\infty)\wedge x\in(0;\infty)}\\[1ex]~=~&(x\,\mathsf e^{-x/z}/z^2)\cdot\mathbf 1_{x\in(0;\infty)\wedge z\in(0;1)}\\[2ex] p_Z(z) ~=~&\int_0^\infty (x\,\mathsf e^{-x/z}/z^2)\cdot\mathbf 1_{z\in(0;1)}\operatorname d x\\[1ex]~=~& \mathbf 1_{z\in(0;1)}\end{align}$$
$\tiny\textrm{Well, okay, the density function turns out to be uniform, but for reasons unrelated to your attempt.}$