On what condition the intersection of all tangent planes of the surface $z = y f(\frac{x}{y})$ is $\{ (0, 0, 0) \}$? Problem
Suppose $f \in C^1$, $A$ is the intersection of all tangent planes of the surface $z = y f(\frac{x}{y})$. Under what condition, we have $A = \{ (0, 0, 0) \}$ ? That is, to find a set $F$, s.t. $A = \{ (0, 0, 0) \} \Leftrightarrow f \in F$.
Analysis
The fact that $(0, 0, 0) \in A$ is discussed sufficently at 1, 2, 3.
I can come up with a trival example to show that there's $f \not\in F$: if $f \equiv 0$, $A = \{ (x, y, 0) | x, y \in \mathbb{R} \}$.
Acknowledge
This is a question derived from a Calculus exercise. But I think answering it may need more difficult math. So I come here to ask for some help.
Thank you!
 A: First, we define$$g(x,y,z)\triangleq z-yf({x\over y})=0$$therefore $$g_x(x,y,z)=-f'({x\over y})\\g_y(x,y,z)=-f({x\over y})+{x\over y}f'({x\over y})\\g_z(x,y,z)=1$$Now, by geometry and calculus and the definition of the gradient vector, we now that and plane tangent on $g(x,y,z)=0$ in $(a,b,c)$ is perpendicular to $\nabla g(a,b,c)$. This leads to the following general equation of planes:$$-f'\left({a\over b}\right)(x-a)+\left[-f\left({a\over b}\right)+{a\over b}f'\left({a\over b}\right)\right](y-b)+z-c=0$$Now we obtain a very interesting fact. After simplification, the general equation, reduces to:$$-f'\left({a\over b}\right)x+\left[-f\left({a\over b}\right)+{a\over b}f'\left({a\over b}\right)\right]y+z=0$$which always crosses the origin. If we define $u\triangleq{a\over b}$, we should go looking for functions, $f$, such that $$\underbrace{\text{Span}}_{u\in \Bbb R}\Bigg[\Big(-f'(u),-f(u)+uf'(u),1\Big)\Bigg]=\Bbb R^3$$Instead, we study cases such that the vector $\Big(-f'(u),-f(u)+uf'(u),1\Big)$ can be imaged by at most 2 independent vectors i.e. $$\Big(-f'(u),-f(u)+uf'(u),1\Big)=k_1(u)(p_1,q_1,r_1)+k_2(u)(p_2,q_2,r_2)$$This requires that a linear combination of the components of the vector be zero, i.e. we can find coefficients $\alpha,\beta,\gamma$ such that$$-\alpha f'(u)+(-f(u)+uf'(u))\beta+\gamma=0$$After solving the outcome differential equation, the possible functions $f(u)$ are $$f(u)=\epsilon u+\zeta$$where $\epsilon ,\zeta$ are real constants. Therefore $$F^c=\{\text{the set of linear functions}\}$$and

$$f(x) \text{ is }\textbf{not} \text{ a linear function}\iff A=\{(0,0,0)\}$$

