$u=2xy=2(r cos⁡θ )(r sin⁡θ )=2r^2 cos⁡θ sin⁡θ=r^2 sin⁡2θ \\ \; \\ \; \\ v=x^2-y^2=r^2 cos^2 θ-r^2 sin^2 θ=r^2 (cos^2 θ-sin^2 θ)=r^2 cos⁡2θ \\ \; \\ \; \\ $

Differentiating u partially w. r. t. r and θ,

$ \therefore \dfrac{\partial u}{\partial r}=2r sin⁡2θ \; ,\; \dfrac{\partial u}{\partial θ}=2r^2 cos⁡2θ \\ \; \\ \; \\ $

Differentiating v partially w. r. t. r and θ,

$ \therefore \dfrac{\partial v}{\partial r} \;=\; 2r cos⁡2θ \; , \; \dfrac{\partial v}{\partial θ} \;=\; =-2r^2 sin⁡2θ \\ \; \\ \; \\ \; \\ \therefore \dfrac{\partial (u,v)}{\partial (r,θ)} \;=\; \left| \begin{array}{cc} \frac{\partial u }{\partial r} & \frac{\partial u }{\partial \theta} \\ \frac{\partial v }{\partial r} & \frac{\partial v }{\partial \theta} \end{array} \right| \;=\; \left| \begin{array}{cc} 2r sin⁡2θ & 2r^2 cos⁡2θ \\ 2r cos⁡2θ & -2r^2 sin⁡2θ \end{array} \right| \\ \; \\ \; \\ \; \\ \therefore \dfrac{\partial (u,v)}{\partial (r,θ)} \;=\; (2r sin⁡2θ )(-2r^2 sin⁡2θ )-(2r^2 cos⁡2θ )(2r cos⁡2θ) \\ \; \\ \; \\ \; \\ \; \\ \therefore \dfrac{\partial (u,v)}{\partial (r,θ)} \;=\; -4r^3 (sin^2 2θ+cos^2 2θ)=-4r^3 (1)=-4r^3 $