BvU said:
Invert the transformation
$$x=r \sin\theta \cos\phi, \quad y=r\sin\theta \sin\phi, \quad z=r\cos\theta$$ to $$r = ...(x, y,z),\quad \phi = ...,\quad \theta = ...$$
(advice: use \sin and \cos, \log etc in ##\LaTeX##)
It should be noted that you
can do this without actually inverting the coordinate relations. The general theory is based on the chain rule. For a function of one parameter only, say ##y(x)##, you would have
$$
1 = \frac{dx}{dx} = \frac{dy}{dx} \frac{dx}{dy}.
$$
This is the reason that ##dx/dy = (dy/dx)^{-1}## in this case (just solve for ##dx/dy##). However, for functions of several variables, say ##y^i(x^1, \ldots, x^n)##, the chain rule would instead give you
$$
\delta^i_j = \frac{\partial y^i}{\partial y^j} = \sum_{k} \frac{\partial y^i}{\partial x^k} \frac{\partial x^k}{\partial y^j}.
$$
Thus, from here you cannot just divide by a factor from the RHS to obtain ##\partial x^k/\partial y^j## since what you have is a sum. However, if you define the matrices ##A## and ##B## defined as
$$
A = \begin{pmatrix}\frac{\partial x^1}{\partial y^1} & \frac{\partial x^1}{\partial y^2} & \ldots \\
\frac{\partial x^2}{\partial y^1} & \frac{\partial x^2}{\partial y^2} & \ldots \\
\vdots & \vdots & \ddots \end{pmatrix}, \qquad
B = \begin{pmatrix}\frac{\partial y^1}{\partial x^1} & \frac{\partial y^1}{\partial x^2} & \ldots \\
\frac{\partial y^2}{\partial x^1} & \frac{\partial y^2}{\partial x^2} & \ldots \\
\vdots & \vdots & \ddots \end{pmatrix},
$$
then the relation can be rewritten
$$
1_n = BA,
$$
where ##1_n## is the ##n \times n## unit matrix. In other words, ##A = B^{-1}##. Thus, if you compute all the derivatives ##\partial y^i/\partial x^j##, you can find ##\partial x^i/\partial y^j## by matrix inversion.
However, if you are unfamiliar with matrices, it is also fine to solve the linear system of equations. For example, in polar coordinates on ##\mathbb R^2##: ##x = r\cos\phi##, ##y = r\sin\phi##, you would find
$$
1 = \frac{\partial x}{\partial x} = \frac{\partial x}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial x}{\partial \phi} \frac{\partial \phi}{\partial x} = \cos\phi \frac{\partial r}{\partial x} - r \sin\phi \frac{\partial \phi}{\partial x}
$$
and
$$
0 = \frac{\partial y}{\partial x} = \frac{\partial y}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial y}{\partial \phi} \frac{\partial \phi}{\partial x} = \sin\phi \frac{\partial r}{\partial x} + r \cos\phi \frac{\partial \phi}{\partial x}
$$
from which you can solve directly for ##\partial r/\partial x## and ##\partial \phi/\partial x##.
.
