## Thomas' Calculus 13th Edition

$1$ and $1$
We know that $f_x(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0+h,y_0)-f(x_0,y_0)}{h}$ $f_x(-2,1)=\lim\limits_{h \to 0} \dfrac{f(-2+h,1)-f(-2,1)}{h}=\lim\limits_{h \to 0} \dfrac{(h-1)-(-1)}{h}=1$ Also, $f_y(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0+h)-f(x_0,y_0)}{h}$ Now, $f_y(-2,1)=\lim\limits_{h \to 0} \dfrac{f(-2,1+h)-f(-2,1)}{h}=\lim\limits_{h \to 0} \dfrac{(2h^2+h-1)-(-1)}{h}=\lim\limits_{h \to 0} (2h+1)=1$