Partial derivatives inverse question If $ \partial u/\partial v=a $, then $ \partial v/\partial u=1/a$?

Functions of a Single Variable
For functions of one variable, if $y=f(x)$ is strictly monotone and differentiable on an interval, and $f'(x)\ne 0$ in that interval, then the inverse function $x=f^{-1}(y)$ is also strictly monotone and differentiable in the corresponding interval and 
$$\bbox[5px,border:2px solid #C0A000]{\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}}\tag 1$$

EXAMPLE:
Suppose $y=\sin(x)$ for $x\in (-\pi/2,\pi,2)$.  Note that the sine function is monotone and differentiable on $(-\pi/2,\pi/2)$ with $\frac{dy}{dx}=\cos(x)$ and $\cos(x)\ne 0$.  
The inverse function, call it $x=\arcsin(y)$ for $y\in (-1,1)$, is  therefore monotone and its derivative is 
$$\frac{dx}{dy}=\frac{1}{\cos(x)}=\frac{1}{\sqrt{1-y^2}}$$
Therefore, we have $\frac{d\,\arcsin(y)}{dy}=\frac{1}{\sqrt{1-y^2}}$.

Functions of a Two Variables
The relationship in $(1)$ does not apply, in general, to functions of more than one variable.  As an example, examine the transformation of Cartesian coordinates $(x,y)$ to polar coordinates $(\rho,\phi)$ as given by 
$$\begin{align}
\rho &=\sqrt{x^2+y^2}\\\\
\phi &=\operatorname{arctan2}(y,x)
\end{align}$$
and
$$\begin{align}
x&=\rho \cos(\phi)\\\\
y&=\rho \sin(\phi)
\end{align}$$
We examine the relationship between $\frac{\partial \rho }{\partial x}$ and $\frac{\partial x}{\partial \rho}$ to see if $(1)$ holds. Note that 
$$\begin{align}
\frac{\partial \rho }{\partial x}&=\frac{x}{\rho}\\\\
& =\cos(\phi)\\\\
&=\frac{\partial x}{\partial \rho}
\end{align}$$
Therefore, $\frac{\partial \rho }{\partial x}\ne \frac{1}{\frac{\partial x}{\partial \rho}}$ and $(1)$ does not hold (unless $y=0$).
Instead of the relationship $(1)$ holding, we have instead
$$\begin{equation}
\begin{pmatrix}
\frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \phi} \\
\frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \phi}
\end{pmatrix}  \begin{pmatrix}
\frac{\partial \rho}{\partial x} & \frac{\partial \rho}{\partial y} \\
\frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y}
\end{pmatrix}=\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\end{equation}$$
whereupon matrix inversion becomes
$$\begin{equation}
\begin{pmatrix}
\frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \phi} \\
\frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \phi}
\end{pmatrix}  =\begin{pmatrix}
\frac{\partial \rho}{\partial x} & \frac{\partial \rho}{\partial y} \\
\frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} 
\end{pmatrix}^{-1} \tag 2\end{equation}$$
Note that $(2)$ is the analog of $(1)$ and applies whenever a transformation and its inverse exists and is prescribed by differentiable functions.  Moreover, it can be generalized to any number of variables.
