Answer
$-13$ and $-2$
Work Step by Step
Since, $f_x(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0+h,y_0)-f(x_0,y_0)}{h}$ and $f_y(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0+h)-f(x_0,y_0)}{h}$
$f_x(1,2)=\lim\limits_{h \to 0} \dfrac{f(1+h,2)-f(1,2)}{h}=\lim\limits_{h \to 0} \dfrac{-6h^2-13h}{h}=-13$
$f_y(1,2)=\lim\limits_{h \to 0} \dfrac{f(1,2+h)-f(1,2)}{h}$
or, $\lim\limits_{h \to 0} \dfrac{-4-2h+4}{h}=-2$
