## Thomas' Calculus 13th Edition

$9$
We know that $f_x(x_0,y_0,z_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0,z_0+h,y_0)-f(x_0,y_0,z_0)}{h}$ $f_x(-1,0,3)=\lim\limits_{h \to 0} \dfrac{f(-1,h, 3)-f(-1,0,3)}{h}=\lim\limits_{h \to 0} \dfrac{(2h^2+9h)-0}{h}=\lim\limits_{h \to 0} (h+9)=9$