Show that $X_1X_2/\sqrt{X_1^2+X_2^2}$ is normally distributed. Where $X_1\sim N(0,\sigma_1^2),X_2\sim N(0,\sigma_2^2)$. Try to use jacobian transformation$$ U=X_1X_2/\sqrt{X_1^2+X_2^2}$$ $$V=X_1$$but I failed to solve the integration.
Edit, $X_1, X_2$ are independent.
Edit2, my effort so far
I use $ U=X_1X_2/\sqrt{X_1^2+X_2^2}$, $V=X_1$ to do the transformation and I solve for $X_1=V,X_2=\frac{UV}{\sqrt{V^2-U^2}}$.
Then I compute $J=\begin{vmatrix} \frac{\partial X_1}{\partial U}&\frac{\partial X_1}{\partial V}\\ \frac{\partial X_2}{\partial U}&\frac{\partial X_2}{\partial V} \end{vmatrix}=\begin{vmatrix}0&1\\-\frac{V^3}{(V^2-U^2)^{\frac{3}{2}}}&\frac{\partial X_2}{\partial V}\\ \end{vmatrix}=\frac{V^3}{(V^2-U^2)^{\frac{3}{2}}}$
Substituting $|J$| and $U,V$ into $f_{X_1,X_2}(x_1,x_2)=f_{X_1}*f_{X_2}$ and integrate with variable $V$
$$f_U(u)=\int_{-\infty}^{+\infty}\frac{1}{2\pi\sigma_1\sigma_2}exp\left(-\frac{v^2}{2\sigma_1^2}-\frac{u^2v^2}{2\sigma_2^2(v^2-u^2)}\right)*|J|dv=\int_{-\infty}^{+\infty}\frac{1}{2\pi\sigma_1\sigma_2}exp\left(-\frac{v^2}{2\sigma_1^2}-\frac{u^2v^2}{2\sigma_2^2(v^2-u^2)}\right)\frac{v^3}{(v^2-u^2)^{\frac{3}{2}}}dv$$
It seems that the integration will get value $0$, since it's symmetric in $v$.
And, I apologize for not showing my effort when I ask a question. I know it's basic manner to do it but I was lazy.
Comment, for intuition and maybe some clues:
I simulated this a million times with with $X_1, X_1 \stackrel{iid}{\sim} Norm(0, 1).$ Then for $Y = X_1X_2/\sqrt{X_1^2 + X_2^2},$ I got $E(Y) \approx 0,$ $SD(Y) \approx 0.5.$ A Shapiro-Wilk test on the first one thousand values of $Y$ failed to reject normality. The histogram of the simulated distribution of $Y$ with the best-fitting normal density function is shown at left below. In this case, $D = X_1^2 + X_2^2 \sim Chisq(df=2)$.
In a second simulation with $X_1 \sim Norm(0,1)$ and independently $X_2 \sim Norm(\mu = 0, \sigma=4),$ I got $E(Y) \approx 0,$ $SD(Y) \approx 0.8.$ A Shapiro-Wilk test on the first one thousand values of $Y$ failed to reject normality. The histogram of the simulated distribution of $Y$ with the best-fitting normal density function is shown at right below.
Perhaps more help if you answer @Did's questions, and show us soon what you tried. What is your Jacobian?