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
We need to take the first partial derivatives of the given function.
In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa:
$f_x=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}} \times \dfrac{\partial (x+yz)}{\partial x}=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}}$
$f_y=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}} \times \dfrac{\partial (x+yz)}{\partial y}=\dfrac{z}{|x+yz| \sqrt {(x+yz)^2-1}}$
$f_z=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}} \times \dfrac{\partial (x+yz)}{\partial z}=\dfrac{y}{|x+yz| \sqrt {(x+yz)^2-1}}$
