Answer
$gradZ(2,3)=4i-j$
Work Step by Step
To find the gradient of $f(x,y)$ denoted $gradf(x,y)$ we use the formula:
$gradf(x,y)=f_{x}(x,y)i+f_{y}(x,y)j$
Note that $f_{x}(x,y)$ and $f_{y}(x,y)$ are partial derivates of the function with respect to $x$ and $y$, respectfully.
$f(x,y)=Z(x,y)=ln(x^2-y)$
$gradZ(x,y)=\frac{2x}{x^2-y}i+\frac{-1}{x^2-y}j$
Substituting in the point $(2,3)$ we have
$gradZ(2,3)=\frac{2(2)}{(2)^2-(3)}i+\frac{-1}{(2)^2-(3)}j$
$gradZ(2,3)=4i-j$