A far more simple method would be to use the gradient.
Lets say we want to get the unit vector $\boldsymbol { \hat e_x } $. What we then do is to take $\boldsymbol { grad(x) } $ or $\boldsymbol { ∇x } $.
This; $\boldsymbol ∇ $, is the nabla-operator. It is a vector containing each partial derivative like this...
$\boldsymbol { ∇= ( \frac {∂} {∂x}, \frac {∂} {∂y}, \frac {∂} {∂z}) } $
When we take the gradient of x we get this...
$\boldsymbol { ∇x= ( \frac {∂x} {∂x}, \frac {∂x} {∂y}, \frac {∂x} {∂z})=(1,0,0)=\hat e_x } $
To get the unit vector of $\boldsymbol x$ in cylindrical coordinate system we have to rewrite $x$ in the form of $\boldsymbol {r_c}$ and $\boldsymbol {\phi}$.
$\boldsymbol {x= r_c cos(x) } $
Now you have to use the more general definition of nabla ($\boldsymbol ∇ $).
Lets say we have a curve-linear coordinate system where the position vector is defined like this...
$\boldsymbol {\vec r = u_1 \hat e_{u1} + u_2 \hat e_{u2} + u_3 \hat e_{u3}} $
... Then the nabla operator for that coordinate system is as follows...
$\boldsymbol { ∇ = \frac {1}{h_1} \frac {∂}{∂u_1} \hat e_{u1} + \frac {1}{h_2} \frac {∂}{∂u_2} \hat e_{u1} + \frac {1}{h_3} \frac {∂}{∂u_3} \hat e_{u1}} $
"$\boldsymbol { h_n } $" is the scale factor to the variable "$\boldsymbol { u_n } $". The scale-factor is defined as: $\boldsymbol {h_n = \frac {\partial \vec r}{\partial u_n}}$
For cylindrical coordinates the position vector is defined as: $\boldsymbol {\vec r = r_c \hat e_{rc} + z \hat e_z }$
With some simple math we can get the scale factors and they are...
$\boldsymbol {h_{rc} = 1 \ \ ,\ h_{\phi} = r_c \ \ ,\ h_z = 1}$
We already know that in cylindircal cooridnates $\boldsymbol x $ is defined as $\boldsymbol {x = r_c cos(x)}$, so now we can get the gradient.
$\boldsymbol { ∇x = ∇(r_c cos(x))= \frac {\partial (r_c \cos(x))}{\partial r_c} \hat e_{rc} + \frac {1}{r_c} \frac {\partial (r_c \cos(x))}{\partial \phi} \hat e_{\phi} + \frac {\partial (r_c cos(x))}{\partial z} \hat e_{z}}$
The result from this gradient is then...
$\boldsymbol {\hat e_{x} = \cos(\phi)\hat e_{rc} - \sin(\phi) \hat e_{\phi}}$
When the same method is applied to $\boldsymbol y$, where $\boldsymbol {y = r_c sin(\phi) }$, we get with ease that...
$\boldsymbol {\hat e_{y} = sin(\phi)\hat e_{rc} + \cos(\phi) \hat e_{\phi}}$
Hope it helped!
I also hope the use of $\boldsymbol \phi $ instead of $\boldsymbol \theta $ and $\boldsymbol {r_c} $ instead of $\boldsymbol \rho $ wasn't to confusing. As a physics student I am more used to the $\boldsymbol {(r_c,\phi,z)}$ standard for cylindrical coordinates.
The coordinate transformation from polar to rectangular coordinates is given by
$$\begin{align}
x&=\rho \cos \phi \tag 1\\\\
y&=\rho \sin \phi \tag 2
\end{align}$$
Now, suppose that the coordinate transformation from Cartesian to polar coordinates as given by
$$\begin{align}
\rho&=\rho (x,y)=\sqrt{x^2+y^2} \\\\
\phi&=\phi(x,y) =
\begin{cases}
\arctan(y/x)&,x>0\\\\
\arctan(y/x)+\pi&,x<0,y\ge 0\\\\
\arctan(y/x)-\pi&,x<0,y<0\\\\
\pi/2&,x=0,y>0\\\\
\pi/2&,x=0,y<0\\\\
\end{cases}
\end{align}$$
were unavailable in closed form. We can still proceed to develop a transformation of the gradient operator from Cartesian coordinates to polar.
To do so, we use the differential relationships
$$\begin{align}
dx&=\frac{\partial x}{\partial \rho}d\rho+\frac{\partial x}{\partial \phi}d\phi\\\\
dy&=\frac{\partial y}{\partial \rho}d\rho+\frac{\partial y}{\partial \phi}d\phi\\\\
d\rho&=\frac{\partial \rho}{\partial x}dx+\frac{\partial \rho}{\partial y}dy\\\\
d\phi&=\frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy
\end{align}$$
Defining the Wronskian $W$ as
$$\begin{align}
W&=\frac{\partial \rho \cos \phi}{\partial \rho}\frac{\partial \rho \sin \phi}{\partial \phi}-\frac{\partial \rho \cos \phi}{\partial \phi}\frac{\partial \rho \sin \phi}{\partial \rho}\\\\
&=\rho
\end{align}$$
we find that
$$\begin{align}
\frac{\partial \rho }{\partial x}&=\frac{1}{W}\frac{\partial \rho \sin \phi}{\partial \phi}\\\\&= \cos \phi\\\\
\frac{\partial \rho }{\partial y}&=-\frac{1}{W}\frac{\partial \rho \cos \phi}{\partial \phi}\\\\&= \sin \phi\\\\
\frac{\partial \phi }{\partial x}&=-\frac{1}{W}\frac{\partial \rho \sin \phi}{\partial \rho}\\\\&=- \frac{\sin \phi}{\rho}\\\\
\frac{\partial \phi }{\partial y}&=\frac{1}{W}\frac{\partial \rho \sin \phi}{\partial \phi}\\\\&=\frac{ \cos \phi}{\rho}
\end{align}$$
Therefore, we have
$$\begin{align}
\hat x \frac{\partial }{\partial x}+\hat y \frac{\partial }{\partial y}&=(\hat \rho \cos \phi -\hat \phi \sin \phi)\left(\frac{\partial \rho}{\partial x}\frac{\partial }{\partial \rho}+\frac{\partial \phi}{\partial x}\frac{\partial }{\partial \phi}\right)+(\hat \rho \sin \phi +\hat \phi \cos \phi)\left(\frac{\partial \rho}{\partial y}\frac{\partial }{\partial \rho}+\frac{\partial \phi}{\partial y}\frac{\partial }{\partial \phi}\right)\\\\
&=\hat \rho \frac{\partial }{\partial \rho}+\hat \phi \frac{1}{\rho}\frac{\partial }{\partial \phi}
\end{align}$$
as was to be shown!
Best Answer
Then $\frac{\partial \phi}{\partial \theta} = 0$. And you can just ignore the middle row of the left transformation matrix, and the middle column of the right transformation matrix.