$g\begin{pmatrix}x\\y \end{pmatrix} = \begin{pmatrix} \sqrt{x^2 + y^2} \\ arctan\frac{y}{x} \end{pmatrix}$
I got $g'\begin{pmatrix}x\\y \end{pmatrix} = \begin{pmatrix} 2x\sqrt{x^2 + y^2} & 2y\sqrt{x^2 + y^2} \\ \frac{1}{1 + y/x^2} & \frac{1}{1 + y^2/x} \end{pmatrix}$
I differentiated $\sqrt{x^2+y^2}$ using the chain rule for both dy and dx.
Everything just looks really weird and I have no intuition whether I did this right or not and would like a second opinion before moving on with the problem, thanks
If $g_1(x,y)=\sqrt{x^2+y^2}=(x^2+y^2)^{1/2}$, then $$\frac{\partial g_1}{\partial x}(x,y)=\frac12(x^2+y^2)^{-1/2}\cdot2x=\frac{x}{\sqrt{x^2+y^2}}$$ the same occurs with respect to $y$. Also, if $g_2(x,y)=\arctan(\tfrac{y}{x})$, then $$\frac{\partial g_2}{\partial x}(x,y)=\cfrac{1}{1+(\tfrac{y}{x})^2}\cdot\Big(-\frac{y}{x^2}\Big)=-\frac{y}{x^2+y^2}$$ and $$\frac{\partial g_2}{\partial y}(x,y)=\cfrac{1}{1+(\tfrac{y}{x})^2}\cdot\frac{1}{x}=\frac{x}{x^2+y^2}$$