Answer
$\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}}$;
$\dfrac{z}{|x+yz| \sqrt {(x+yz)^2-1}}$;
$\dfrac{y}{|x+yz| \sqrt {(x+yz)^2-1}}$
Work Step by Step
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by treating $y$ and $z$ as a constant:
$f_x=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}} \times \dfrac{\partial }{\partial x} (x+yz)=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}}$
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $y$, by treating $x$ and $z$ as a constant:
$f_y=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}} \times \dfrac{\partial }{\partial y} (x+yz)=\dfrac{z}{|x+yz| \sqrt {(x+yz)^2-1}}$
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $z$, by treating $x$ and $y$ as a constant:
$f_z=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}} \times \dfrac{\partial }{\partial z} (x+yz)=\dfrac{y}{|x+yz| \sqrt {(x+yz)^2-1}}$
