Proving there is no plane tangent to a graph Define $$f(x, y) = \begin{cases} 
\sin(y^2/x)\sqrt{x^2 + y^2} & \text{ if } x \neq 0 \\
0 & \text{ if } x = 0.
\end{cases}$$
(a) Show $f : \mathbb{R}^{2} \rightarrow \mathbb{R}$ has directional derivatives in every direction at $(0, 0)$.
(b) Show there is no plane that is tangent to the graph of $f : \mathbb{R}^{2} \rightarrow \mathbb{R}$ at the point $(0, 0, f(0,0))$.

Geometrically, part $(b)$ makes sense because it's a really sharp point (but still continuous), meaning that there should be no tangent plane there. I don't fully understand how I'm supposed to show there's no tangent at $(0, 0, f(0,0))$ though. Isn't this a point in $\mathbb{R}^{3}$? And we have a function in $\mathbb{R}^{2}$?
I'm also not too sure about how to do $(a)$. I know how to compute the directional derivatives. I don't think that's enough to show existence, though. I'm guessing that it's going to be some sort of limit.
I would really appreciate some sort of help with this exercise. 
 A: Part (a) is fundamentally a calculation. The directional derivative in the $v$ direction (for a unit vector $v$) at zero is defined as $\lim_{t\to 0}\frac{f(tv)-f(0)}{t}$. So, what is that here? Let $v=(a,b)$ with $a^2+b^2=1$; then $f(tv)=\sin\frac{t^2b^2}{ta}\cdot\sqrt{t^2a^2+t^2b^2}=\sin\left(t\frac{b^2}{a}\right)\cdot t$. Divide by $t$, and we get $\sin \left(t\frac{b^2}{a}\right)$, which goes to zero as $t\to 0$ for fixed nonzero $a$ and fixed $b$. The directional derivatives exist in all directions are all zero. Well, OK, we have to treat the directional derivative in the $(0,1)$ direction as a special case; that one uses the other formula for $f$.
Part (b) deserves an explanation that goes more into the theory.
A function $f$ from $\mathbb{R}^2$ to $\mathbb{R}$ is differentiable at a point $u_0=(x_0,y_0)$ if and only if there is a linear map $Df$ from $\mathbb{R}^2$ to $\mathbb{R}$ such that
$$\lim_{\|v\|\to 0}\frac{f(u_0+v)-f(u_0)-Df(v)}{\|v\|}=0$$
The extra dimensions mean we can't just divide $f(u_0+v)-f(u_0)$ by $v$, so we substitute with a linear map. Now, this linear map also defines a plane: the function $g(x,y)=f(x_0,y_0)+Df(x-x_0,y-y_0)$ is the equation of a plane, and we call this the tangent plane.
That is the definition of differentiability, and of a tangent plane. If a function isn't differentiable, then no linear map allows for that plane to be close enough to the graph of the function. So then, what this part needs us to show is that the function isn't differentiable. Note that the prior calculation with the directional derivatives doesn't suffice here; it's enough to tell us that if the function is differentiable, its derivative is zero - but only that.
So, based on the hypothesized derivative of zero, and the fact that $f(0)=0$, we need to show that $\frac{f(v)}{\|v\|}$ doesn't go to zero as $v\to 0$. Well, looking at the formula for $f$, dividing by the norm $\|v\|=\sqrt{x^2+y^2}$ just leaves us with the $\sin\frac{y^2}{x}$ term. Can we find points close to the origin where that $\sin$ is far from zero?
A: (a): Let $v=(v_1,v_2)$ be a direction.
Then show that $ \lim_{t \to 0}\frac{f(tv)-f(0,0)}{t}$ exists and $=0$.
(b): from (a) we get $f_x(0,0)=0=f_y(0,0)$. 
In general the plane that is tangent to the graph at $(0,0,f(0,0)$ is given by
$z=f(0,0)+f_x(0,0)x+f_y(0,0)y.$
But in our case we have $f(0,0)=0$ and $f_x(0,0)=0=f_y(0,0)$. 
Conclusion ?
