Let $I(x)=\sigma(x)/x$ be the abundancy index of the positive integer $x$, where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.
Consider $$f(q,k)=I(q^k)+\frac{2}{I(q^k)}$$ where $5 \leq q$ is a prime number and $k$ is a positive integer.
We obtain the partial derivatives $$\frac{\partial f}{\partial k} = -\frac{\bigl(q^{2k+2} - 4q^{2k+1} + 2q^{2k} + 2q^{k+1} - 1\bigr)\ln q}{q^k (q - 1)(q^{k+1} - 1)^2} < 0$$ and $$\frac{\partial f}{\partial q} = \frac{\bigl(q^{k+1} - (k+1)q + k\bigr)\bigl(q^{2k+2} - 4q^{2k+1} + 2q^{2k} + 2q^{k+1} - 1\bigr)}{q^{k+1} (q - 1)^2 (q^{k+1} - 1)^2} > 0.$$
We then get the partial differential equation $$\frac{q(q - 1)}{q^{k+1} - (k+1)q + k}\cdot\frac{\partial f}{\partial q}=-\frac{1}{\ln q}\cdot\frac{\partial f}{\partial k}=\frac{q^{2k+2} - 4q^{2k+1} + 2q^{2k} + 2q^{k+1} - 1}{q^k (q - 1) (q^{k+1} - 1)^2}.$$
Here then is my question, for which I do not know a satisfactory answer yet, as I am not an expert on PDEs:
Is there a function $g(q,k) \neq f(q,k)$ such that $$\frac{q(q - 1)}{q^{k+1} - (k+1)q + k}\cdot\frac{\partial g}{\partial q}=-\frac{1}{\ln q}\cdot\frac{\partial g}{\partial k}?$$ Of course, I would like to specifically exclude $$\frac{\partial g}{\partial q}=\frac{\partial g}{\partial k}=0.$$
Posting this self-answer, so that this question does not remain in the unanswered queue.
Let $$g(h,k) = f(h,k) + c,$$ for some absolute constant $c \neq 0$. (Note then that $g(h,k) \neq f(h,k)$ holds.)
Then $g(h,k)$ satisfies the constraints of our problem, since we obtain the partial derivatives $$\frac{\partial g}{\partial k} = \frac{\partial f}{\partial k} = -\frac{\bigl(q^{2k+2} - 4q^{2k+1} + 2q^{2k} + 2q^{k+1} - 1\bigr)\ln q}{q^k (q - 1)(q^{k+1} - 1)^2} < 0$$ and $$\frac{\partial g}{\partial q} = \frac{\partial f}{\partial q} = \frac{\bigl(q^{k+1} - (k+1)q + k\bigr)\bigl(q^{2k+2} - 4q^{2k+1} + 2q^{2k} + 2q^{k+1} - 1\bigr)}{q^{k+1} (q - 1)^2 (q^{k+1} - 1)^2} > 0.$$
We then get the partial differential equation $$\frac{q(q - 1)}{q^{k+1} - (k+1)q + k}\cdot\frac{\partial g}{\partial q}=-\frac{1}{\ln q}\cdot\frac{\partial g}{\partial k}=\frac{q^{2k+2} - 4q^{2k+1} + 2q^{2k} + 2q^{k+1} - 1}{q^k (q - 1) (q^{k+1} - 1)^2}.$$