Suppose that $f(x,y)$ is a continuously differentiable function and $g(x,y) =xy-f(x,y)$. I know that $g$ is concave if and only if $(-f_{xx})(-f_{yy}) -(1-f_{xy}) ^{2}>0$ and $f_{xx}>0$.
Now suppose that I "travel" along the function $g$ on a path that satisfies $f_{x}=y$. Thus, along this path $f_{xy}=1$.
Is it correct that along the path the function $g$ is concave if and only if $ (-f_{xx}) (-f_{yy}) >0$ and $f_{xx}>0$ ? An explanation is much appreciated.
Of course, your "if and only if" statement is incorrect. E.g., let $$f(x,y):=x^2/2+y^2/4.$$ Then your conditions $(-f_{xx})(-f_{yy})>0$ and $f_{xx}>0$ obviously hold. Further, here $f_x(x,y)=y$ means $y=x$, and hence the concavity of $g$ along the path means that $g(x,x)$ is concave in $x$. However, $g(x,x)=x^2/4$ is strictly convex, and thus not concave, in $x$.
As for your reasoning, if you try to formalize it, avoiding terms such as "along the path" and "travel", perhaps you will see where you made mistakes. Hints: (i) $f_x(x,y(x))=y(x)$ does not imply $f_{xy}(x,y(x))=1$ (you are confusing independent and dependent variables here) and (ii) the concavity of $g(x,y)$ in $(x,y)$ means that the restriction of $g$ to every straight line is concave, and this concavity does not in general imply any kind of concavity over curved lines.
