Answer
$-2$
Work Step by Step
Our aim is to take the first partial derivative of the given function $f(x,y,z)$ with respect to $x$, by treating $y$ and $z$ as a constant:
$f_x= \dfrac{\partial (xy+z^3 x-2yz)}{\partial x}=0$
or, $ y+3xz \dfrac{\partial (z)}{\partial x}+z^3-2y\dfrac{\partial (z)}{\partial x}=0$
$f_x(1,1,1)=y+3xz \dfrac{\partial (z)}{\partial x}+z^3-2y\dfrac{\partial (z)}{\partial x}$
or, $ 1+3(1)(1) \dfrac{\partial (z)}{\partial x}+(1)^3-2 (1)\dfrac{\partial (z)}{\partial x}=0$
or, $\dfrac{\partial (z)}{\partial x}=-2$
