Understanding secular equation derivation
It doesn't matter how you obtain an expression for the partial derivative, as long as you obtain one, and set the partial derivative to zero. The method you described above is probably easier than differentiating the fraction. All the steps that you did are mathematically valid, so your final result is valid. (The "differentiate both sides" step is valid because the equality holds for all values of $c_a$ and $c_b$.)

If you want to, you can differentiate the fraction and you will end up with the same result (i.e. the secular equations). I actually typed the maths out before, so here you go: https://chemistry.stackexchange.com/a/42636/16683 The notation is a tiny bit different, but you should be able to understand it.



As for the linear algebra parallel, the secular equations are only analogous to $Ax = \lambda x$ if you make the additional assumption that $S_{ab} = 0$, in which case $H-ES$ is directly analogous to $A - \lambda I$ (since the overlap matrix is now equivalent to the identity matrix). In this case, the secular equations reduce to $Hc = Ec$.

If you do not make that assumption, then the secular equations take the form of $Hc = ESc$ and the parallel you should draw is to the generalised eigenvalue equation $Ax = \lambda Bx$. The eigenvectors $x$ are said to be the "eigenvectors of $A$ with respect to $B$". There is a lot more information online if you are interested - simply google "generalised eigenvalue".