Answer
$\lt 2xye^{\frac{y}{z}}, x^{2}(e^{\frac{y}{z}}+\frac{e^{\frac{y}{z}}y}{z}), -\frac{x^{2}y^{2}e^{\frac{y}{z}}}{z^2}\gt$
Work Step by Step
Using the chain rule of differentiation we have:
$f_{x}(x, y, z) = 2xye^{\frac{y}{z}}$
$f_{y}(x, y, z) = x^{2}(e^{\frac{y}{z}}+\frac{e^{\frac{y}{z}}y}{z})$
$f_{z}(x, y, z) = -\frac{x^{2}y^{2}e^{\frac{y}{z}}}{z^2}$
Therefore, the gradient vector of $f$ is:
$\lt 2xye^{\frac{y}{z}}, x^{2}(e^{\frac{y}{z}}+\frac{e^{\frac{y}{z}}y}{z}), -\frac{x^{2}y^{2}e^{\frac{y}{z}}}{z^2}\gt$
