Show that the graph $z = f(x,y)$ that intersects every vertical plane $y = \lambda x$ in a straight line is a linear function Let $f:\mathbb{R}^{2} \to \mathbb{R}$ be a differentiable function whose graph $z = f(x,y)$ intersects every vertical plane $y = \lambda x$ in a straight line.
Show that $f$ is a linear function $f(x,y) = ax + by + c$
I'm not sure what it is, perhaps I am really overthinking things, but I am struggling to work this question out.  Perhaps it's because I'm not sure what is meant by linear function in this question, but I am given the exact form that the function should be in, which to me means that the function is linear in all of its variables. This being said I am still stuck.
What  came to mind would be to perhaps use some sort of tangent plane, but that doesn't seem to me like the right thing to do...
EDIT: There does appear to have been a solution posted ( A differentiable function intersecting every vertical plane $y = \lambda x$ must be linear ) and the question is a bit harder than I first imagined, but there is no context to how the steps are taken. To start right from the beginning I don't understand:
1) How the function $\phi(\lambda)$ came to be existing.
2) What brought about defining or knowing about $f(0,0)$.
3) How the first difference quotient defined even came about.
And I struggled with the rest of the solution onward.
Would someone be willing to explain things a bit more clearly? 
 A: Geometrically the given hypothesis implies that the curvature of the surface must be zero at the intersecting curve.
This implies $z=f(x,y=\lambda x)=c$ or $dz=0$.
$\implies \frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy=0$
$\implies (p+\lambda q)dx=0$ where $p=\partial z/\partial x$ and $q=\partial z/\partial y$. 
Picking the partial differential equation $(p+\lambda q)=0$.
This gives $p=-\lambda q=k$(say)
Using $dz=pdx+qdy$ you have $dz=kdx-(k/\lambda)dy$
Integrate, $z=kx-(k/\lambda) y+ c$ is solution.
A: WLOG, $f(0,0)=0.$ I'll assume the graph of $f$ over every line through the origin is itself a line.
Let $v$ be a nonzero vector in $\mathbb R^2.$ For  $t\in \mathbb R,$ define $f_v(t) = f(tv).$ Our assumption on $f$ shows $f_v(t) = Ct$ for some constant $C.$ Here we've used $f(0,0)=0.$
Now $C= (f_v)'(0).$ Because $f$ is differentiable at $(0,0),$ we can use the chain rule to calculate this. We get $C=Df(0,0)(v).$ Thus
$$f(tv)= f_v(t) = Df(0,0)(v)\cdot t = Df(0,0)(tv).$$
This says $f = Df(0,0)$ on the line $\{tv\}.$ Since $v$ was arbitrary, $f = Df(0,0)$ on every line through the origin. Thus $f = Df(0,0)$ on $\mathbb R^2.$ In other words,
$$f(x,y)=\frac{\partial f}{\partial x}(0,0) x + \frac{\partial f}{\partial y}(0,0)y,\,\, (x,y)\in \mathbb R^2.$$
