This week's problem was correctly answered by MarkFL. You can find his solution below.
[sp]We are given:
$$z(x,y)=xf\left(\frac{y}{x} \right)$$
Let us the define:
$$F(x,y,z)=xf\left(\frac{y}{x} \right)-z$$
Thus, a point $\left(x_0,y_0,z_0 \right)$ is on the graph of $$z(x,y)=xf\left(\frac{y}{x} \right)$$ iff it is also on the level surface $F(x,y,z)=0$. This follows from $$F\left(x_0,y_0,z_0 \right)=x_0f\left(\frac{y_0}{x_0} \right)-z_0=0$$.
Next, we may utilize the following theorem:
Equation of Tangent Plane
Let $P\left(x_0,y_0,z_0 \right)$ be a point on the graph of $F(x,y,z)=c$ where $\nabla F\ne0$. Then an equation of the tangent plane at $P$ is:
$$F_x\left(x_0,y_0,z_0 \right)\left(x-x_0 \right)+F_y\left(x_0,y_0,z_0 \right)\left(y-y_0 \right)+F_z\left(x_0,y_0,z_0 \right)\left(z-z_0 \right)=0$$
Computing the required partials, we find:
$$F_x(x,y,z)=xf'\left(\frac{y}{x} \right)\left(-\frac{y}{x^2} \right)+f\left(\frac{y}{x} \right)=f\left(\frac{y}{x} \right)-\frac{y}{x}f'\left(\frac{y}{x} \right)$$
$$F_y(x,y,z)=xf'\left(\frac{y}{x} \right)\left(\frac{1}{x} \right)=f'\left(\frac{y}{x} \right)$$
$$F_z(x,y,z)=-1$$
Using the above theorem, we then find the equation of the tangent plant at $P$ is:
$$\left(f\left(\frac{y_0}{x_0} \right)-\frac{y_0}{x_0}f'\left(\frac{y_0}{x_0} \right) \right)\left(x-x_0 \right)+\left(f'\left(\frac{y_0}{x_0} \right) \right)\left(y-y_0 \right)+(-1)\left(z-z_0 \right)=0$$
Distributing and rearranging, we obtain:
$$]\left(f\left(\frac{y_0}{x_0} \right)-\frac{y_0}{x_0}f'\left(\frac{y_0}{x_0} \right) \right)x+\left(f'\left(\frac{y_0}{x_0} \right) \right)y-z-\left(x_0f\left(\frac{y_0}{x_0} \right)-z_0 \right)=0$$
Now, given $$F\left(x_0,y_0,z_0 \right)=x_0f\left(\frac{y_0}{x_0} \right)-z_0=0$$, this reduces to:
$$]\left(f\left(\frac{y_0}{x_0} \right)-\frac{y_0}{x_0}f'\left(\frac{y_0}{x_0} \right) \right)x+\left(f'\left(\frac{y_0}{x_0} \right) \right)y-z=0$$
Observing that all planes given by $ax+by+cz=0$ pass through the origin, we therefore find that all of the planes tangent to $z$ must have the origin as a common point. [/sp]