So I'm kinda new to partial derivatives and I'm supposed to find the tangent plane for $z=yf\left ( \frac{x}{y} \right )$ but I'm kinda thrown off by the $f\left ( \frac{x}{y} \right )$ part when it comes to finding the partial derivatives.
$\frac{\partial z }{\partial x} = 0\cdot f\left ( \frac{x}{y} \right )+ \left ( \frac{\partial f }{\partial x} \cdot \frac{1}{y} \right )\cdot y = \frac{\partial f }{\partial x}$
$\frac{\partial z }{\partial y} = 1\cdot f\left ( \frac{x}{y} \right )+ \left ( \frac{\partial f }{\partial y} \cdot \frac{-x}{y^2} \right )\cdot y = \frac{\partial f }{\partial y}\cdot \frac{-x}{y}+f\left ( \frac{x}{y} \right )$
is what I'm getting but I'm not sure whether it's right.
One nitpick. The function $f$ is a function of a single variable, so $\partial f/\partial x$ and $\partial f/\partial y$ do not make sense here -- only $f'$. So if $z = yf(x/y)$, we have $$\frac{\partial z}{\partial x}(x,y) = f'\left(\frac{x}{y}\right) \quad \mbox{and} \quad \frac{\partial z}{\partial y}(x,y) = f\left(\frac{x}{y}\right) - \frac{x}{y}f'\left(\frac{x}{y}\right).$$