Verify $JJ' = 1$, given $x = e^u \cos v$, $y = e^u \sin v$. 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:
$$
f(x,y) = y/x = y \cdot x^{-1} \implies 
\frac{\partial f}{\partial x} = -y x^{-2} = -y \cdot \frac{1}{x^2} = \frac{-y}{x^2}.
$$
