A Double Partial Derivative and Integral Consider the following differential equation
$$\frac{\partial z}{\partial \phi} . \frac{\partial z }{\partial \mu} = 1$$
My attempt at finding the solution for the complete integral of this is
$$\int \int dz\,dz = \int \int d \mu \,d\phi$$
$$\int z\,dz = \int (\mu +A) \, d\phi$$
$$\frac{z^2}{2} = \mu \phi  + A \phi + B$$
$$z = \sqrt{2\phi\mu +2A\phi +2B}$$
where $A$ and $B$ are arbitrary constants. However, referring to my professor's solution, the family of solutions should be given by
$$z=A\phi +\frac{1}{A}\mu + B$$
which is obviously not in agreement with my own findings. Can anyone see where I am going wrong?
 A: The differential equation is equivalent to
$$\frac{\partial z}{\partial \phi} = \left(\frac{\partial z}{\partial \mu}\right)^{-1},$$
which gives us a hint that we should assume a separable form of $z$, such as
$$z = f(\phi) + g(\mu) + B,$$
where $f$ is only a function of $\phi$, $g$ is only a function of $\mu$, and $B$ is a constant to be determined by initial conditions on $z$. Note that other separable forms of $z$ can also be assumed (as explained below). Given this form of $z$, the above differential equation becomes something much easier to manage:
\begin{align}
\tag{1}
\frac{\partial f}{\partial \phi} = \left(\frac{\partial g}{\partial \mu}\right)^{-1}.
\end{align}
Here, we've dropped the arguments of each function for brevity of notation. From this last equation, we can choose a convenient expression for $f$, so that the expression for $g$ will be simple. For example, by choosing $f(\phi) = A\phi$, a simple choice for $g$ would be $g(\mu) = \mu/A$, so that
$$\left(\frac{\partial g}{\partial \mu}\right)^{-1} = \left(\frac{1}{A}\right)^{-1} = A = \frac{\partial f}{\partial \phi},$$
and the differential equation is satisfied. Hence, a solution (family) is
$$z = A\phi + \frac{1}{A}\mu + B.$$
Remark. It turns out that there are (practically) no other possible choices of $f$ and $g$. In equation $(1)$, the partial derivative on the left can't contain $\mu$ since $f$ is not a function of $\mu$, and similarly, the partial derivative on the right can't contain $\phi$. So, neither side of the equation can contain $\phi$ nor $\mu$. The only way for the equation to hold is if $f = a\phi + b$ for some constants $a$ and $b$, so that its partial derivative contains neither $\phi$ nor $\mu$. Likewise $g(\mu) = c\mu + d$ for some constants $c$ and $d$. Plugging these expressions into the assumed separable form of $z$, we notice that $b$ and $d$ would get absorbed into $B$, and we'd then infer (as we just did above) that the remaining constants are $a = 1/c$ (which gives the same answer as before, by using the labels $a = A$ and $c = 1/A$).

Above, we assumed that $z$ had an additively separable form, but we could have also assumed a multiplicatively separable form:
$$z = f(\phi)g(\mu) + B.$$
After some rearrangement, the differential equation would then become
$$f\frac{\partial f}{\partial \phi}\left(g\frac{\partial g}{\partial \mu}\right) = 1,$$
which gives us a hint as to what choices of $f$ and $g$ are convenient. For example, if we choose $f(\phi) = \sqrt{2\phi}$ and $g(\mu) = \sqrt{2\mu}$, then 
$$\frac{\partial f}{\partial \phi} = \frac{1}{\sqrt{2\phi}} = \frac{1}{f},$$
and likewise, $\partial g/\partial \mu = 1/g$. Then the differential equation is satisfied, and hence,
$$z = 2\sqrt{\phi\mu} + B$$ is also a solution (family). Were there other possible choices of $f$ and $g$ here? I'll leave it up to the interested reader to investigate.

By the way, your approach using direct integration doesn't work as straightforwardly as you might think. Even if you're given an "isolated" partial derivative, you must understand its integral in the following sense:
https://math.stackexchange.com/a/606708/436135.
So if you're given the equation $\partial z/ \partial \phi = 1$, it doesn't imply that $dz = d\phi$ because these symbols are meaningless in this context. You may work with these symbols, if you wish, by giving them proper meaning as differentials:
https://en.wikipedia.org/wiki/Differential_of_a_function.
But it doesn't seem useful to do this for the problem at hand. In fact, if you have a justifiable need for working with differentials, you'd certainly know it, and in all other cases, you'd be better off avoiding it. The interpretation of derivatives as ratios of differentials has a history, which is discussed here:
https://en.wikipedia.org/wiki/Leibniz%27s_notation.
As you can see, the notion of differential has evolved over the centuries. Its modern usage is varied and significant. In all situations, however, its underlying machinery is more sophisticated than the way in which you attempted to use it.
