Concavity of a function along a path 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.

You are incorrect that $f_{xy}=1$ along the path $y=f_x$. For example, consider $f(x,y)=x^2+y^2$. Then $f_{xy} = 0$, whether I move along the path $y=2x$ or not.
What you want to do is consider a curve $\phi(t)=(x(t),y(t))$ along which $y=f_x$. Set $G(t)=g(x(t),y(t))$. You're inquiring about $G''(t)$. By a chain rule computation, you will get
\begin{align*}
G''(t) &= g_{xx}x'(t)^2 + 2g_{xy}x'(t)y'(t)+g_{yy}y'(t)^2 + g_x x''(t) + g_y y''(t) \\
&= -f_{xx}x'(t)^2 + 2(1-f_{xy})x'(t)y'(t) - f_{yy}y'(t)^2 + (y-f_x)x''(t) + (x-f_y)y''(t).\end{align*}
(I've been sloppy with indicating functions of $t$. Of course, all partials, as well as $x$ and $y$, are evaluated at $(x(t),y(t))$.)
Now, the equation you gave for the curve tells us that
$$y'(t) = f_{xx}x'(t) + f_{xy}y'(t), \quad\text{or}\quad f_{xx}x'(t)+\big(f_{xy}-1\big)y'(t)=0.$$
I will let you finish the algebra of substituting and (with hope) simplifying to see what you get.
