Answer
$9$
Work Step by Step
$f_x(-1,0,3)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0,z_0+h,y_0)-f(x_0,y_0,z_0)}{h}=\lim\limits_{h \to 0} \dfrac{f(-1,h, 3)-f(-1,0,3)}{h}$
or, $=\lim\limits_{h \to 0} \dfrac{(2h^2+9h)-0}{h}$
or, $\lim\limits_{h \to 0} (h+9)=9$
