Answer
$12$
Work Step by Step
We know that $f_x(x_0,y_0,z_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0,z_0+h,y_0)-f(x_0,y_0,z_0)}{h}$
$f_x(1,2,3)=\lim\limits_{h \to 0} \dfrac{f(1,2,3+h)-f(1,2,3)}{h}=\lim\limits_{h \to 0} \dfrac{(18+12h+h^2)-18}{h}=\lim\limits_{h \to 0} (12+2h)=12$