I was trying to solve this jacobian problem but I'm not getting the required solution i.e $JJ' = 1$
I'm getting $J = e^{2u}$ and $J' = \frac{x^2 - y^2}{x^2 + y^2}$ So multiplying both $J*J'$ will not return 1 in my case.
What am I doing wrong? Here's my solution,
You're almost there. First, a miscalculation: your $\frac{\partial v}{\partial x}$ has the wrong sign. In particular: we indeed have $$ u = \frac 12 \log(x^2 + y^2), \quad v = \tan^{-1}(y/x). $$ It follows that $$ \frac{\partial u}{\partial x} = \frac{x}{x^2 + y^2}, \quad \frac{\partial u}{\partial y} = \frac{y}{x^2 + y^2}, $$ which you calculated correctly. However,
$$ \frac{\partial v}{\partial x} = \frac{-y}{x^2} \cdot \frac{1}{1 + (y/x)^2} = \frac{-y}{x^2 + y^2}, \qquad \frac{\partial v}{\partial y} = \frac{1}{x} \cdot \frac{1}{1 + (y/x)^2} = \frac{x}{x^2 + y^2}. $$
Now, in order to see that $JJ' = 1$, write both Jacobians as a function of the same set of variables. That is, either as a function of $x$ and $y$ or as a function of $u$ and $v$.
If we write $J'$ as a function of $u$ and $v$, then we find $$ J' = \frac{x^2 + y^2}{(x^2 + y^2)^2} = \frac{1}{x^2 + y^2} = \frac{1}{[e^u \cos v]^2 + [e^u \sin v]^2} = \frac{1}{e^{2u}}. $$
In greater detail: we have $v(x,y) = \tan^{-1}(y/x)$. We find the derivative of this using the chain rule. Because $\frac{d}{dt}\tan^{-1}(t) = \frac{1}{1 + t^2}$, we have $$ \frac{\partial v}{\partial x} = \frac{1}{1 + (y/x)^2} \cdot \frac{\partial}{\partial x}[(y/x)]. $$ To calculate the partial derivative of $y/x$ with respect to $x$, we use the power rule. In particular: $$ f(x,y) = y/x = y \cdot x^{-1} \implies \frac{\partial f}{\partial x} = -y x^{-2} = -y \cdot \frac{1}{x^2} = \frac{-y}{x^2}. $$