finding the values of a such that an implicit function g(y)=x has max,min,saddle points along y=0

Question

I've got

$$f(x,y)= a\exp(1+xy) + a^2 \sin(x) +1$$

for which I've shown that there exists an implicit function $x=g(y). ( df/dx \neq 0)$

and $df/dx = a y \exp(1+xy) + a^2 \cos x$ now in the neighborhood of $P=(0,0)$ for the implicit function to exist I'd need $a y \exp(1+xy) \neq 0$ but at $P$, wouldn't this be 0?

Given, $g(y)=x$, how do I find the max,min,saddle points?

Answer 1

This is not a complete answer just some work I was doing using the second derivative test to locate the maxima, minima and saddle points.

$\frac{\partial f}{\partial x} = y \ln a (a^{1+xy}) + a^2\cos x$

$\frac{\partial f}{\partial y} = x \ln a (a^{1+xy})$

Now we must set these equal to zero. Setting the y partial equal to zero gives a solution of (0,y) but since we want y = 0 we have for testing the point (0,0). However, for this to be a maximin, it must also be a valid point in the other equation set equal to zero, which it isn't unless a too equals zero. However, ln 0 is undefined so...perhaps there is no such a? Thoughts?