Answer
$\frac{\partial z}{\partial x}\frac{\partial^2 z}{\partial x \partial t}=\frac{\partial z}{\partial t} \frac{\partial^2 z}{\partial x^2}$
Work Step by Step
Given: $z=sin(x+sint)$
$\frac{\partial z}{\partial x}=cos(x+sint)$
$\frac{\partial z}{\partial t}=cos(x+sint) \cdot cost$
$\frac{\partial^2 z}{\partial x \partial t}=-sin(x+sint) \cdot cost$
$\frac{\partial^2 z}{\partial x^2}=-sin(x+sint)$
Hence, $\frac{\partial z}{\partial x}\frac{\partial^2 z}{\partial x \partial t}=\frac{\partial z}{\partial t} \frac{\partial^2 z}{\partial x^2}$
