Can someone clarify the relationship between directional derivative and partial derivatives for a function from $\mathbb{R}^n$ to $\mathbb{R}$?
To my understanding, if the function is continuously differentiable, then both directional and partial derivatives exist. Is that correct?
Consider this function: \begin{align} f(x,y) = \begin{cases} \sin( \frac{y^2}{x})\sqrt{x^2 + y^2}& x \ne 0 \\ 0 & x = 0 \end{cases} \end{align} How would one verify this function has direction derivatives at $(0,0)$?
To be short:
One could say partial derivatives are special directional derivatives. Assuming existence, one can use partial derivatives to calculate directional derivativs.
Suppose $E$ is an open subset of $\mathbb{R}^n$. If $f:E\to\mathbb{R}^m$ is continuously differentiable, then all the partial derivatives exist and are continuous on $E$. The converse is also true. See for instance Theorem 9.21 in Rudin's Principle of Mathematical Analysis.
Regarding your example: \begin{align} f(x,y) = \begin{cases} \sin( \frac{y^2}{x})\sqrt{x^2 + y^2}& x \ne 0 \\ 0 & x = 0 \end{cases} \end{align} "How would one verify this function has direction derivatives at $(0,0)$?"
One needs to specify the direction. For instance in the direction of $u=\frac{1}{\sqrt2}(1,1)$, one has $$ f(x,x)=\sqrt 2 x\sin x,\quad x\geq 0. $$ Thus the functional derivative at $(0,0), $in the direction $u$ exists. You can similarly check other directions by letting $y=mx$ and $x=ny$.
Rudin explains clearly in his book partial derivatives, directional derivatives and their relation:
Let $f: \mathbb{R}^n \to \mathbb{R}$ and $x^1,...,x^n$ be the standard coordinate functions on $\mathbb{R}^n$. Suppose $f$ is also differentiable about $p$. Then we define the derivative of $f$ in the direction $\vec{v}$ at a point $p$ to be:
$$D_{\vec{v}} f(p) = \lim_{t \to 0} \frac{ f(p+t\vec{v})-f(p)}{t} = \nabla f(p) \cdot \vec{v} = \sum_j v^i\frac{\partial f}{\partial x^j}(p)$$
Observe that if $\vec{v} = \textbf{e}^i = \langle 0,...,x^i=1,...,0\rangle$ then the above definition becomes:
$$D_{\textbf{e}^i}f(p) = \frac{\partial f}{\partial x^i}(p)$$
i.e directional derivatives are a generalization of partial derivatives. If you wish to compute the partials at $(0,0)$ for your function, you will have to proceed by definition.
$$\frac{\partial f}{\partial x}(0,0) = \lim_{t = 0} \frac{f(t,0) - f(0,0)}{t} = \lim_{t \to 0} \frac{f(t,0)-0}{t} = 0$$
$$\hspace{-.4in} \frac{\partial f}{\partial y}(0,0) = \lim_{t = 0} \frac{f(0,t) - f(0,0)}{t} = \lim_{t \to 0} \frac{0-0}{t} = 0$$
Hint: To check whether all directional derivatives at $(0,0)$ exist, all we need to is answer (i) Does $x\to f(x,mx)$ have a one-variable derivative at $x=0$ for any $m\in \mathbb R$? (ii) Does $y\to f(0,y)$ have a one-variable derivative at $y=0?$ These are not to hard to answer here; give it a shot.
Let me add to the discussion. Consider the following example \begin{align} f(x,y) = \begin{cases} \frac{x^2-y^2}{x^2+y^2} &\ \text{ if } \ \ (x,y) \neq (0, 0)\\ 0 & \ \text{ if } \ \ (x, y) = (0 , 0) \end{cases} \end{align} which is not continuous at $(0, 0)$.
It's doesn't have partial derivatives at $(0, 0)$ since \begin{align} \lim_{h\rightarrow 0} \frac{f(h, 0)-f(0, 0)}{h} = \lim_{h\rightarrow 0}\frac{1}{h} \end{align} which doesn't exists and likewise \begin{align} \lim_{k\rightarrow 0} \frac{f(0, k)-f(0, 0)}{k} = \lim_{k\rightarrow 0} \frac{-1}{k}. \end{align} Nevertheless, if we look at the direction $x=y$, we have \begin{align} \lim_{t\rightarrow 0} \frac{f(t, t)-f(0, 0)}{t}= 0 \end{align} that is $f$ has a directional derivative in the $x=y$ direction and the directional derivative is definitely not given by \begin{align} \nabla f(0,0)\cdot (1/\sqrt{2}, 1/\sqrt{2}). \end{align}
Here's a hint for your problem. Consider your given function, then \begin{align} \lim_{t\rightarrow 0} \frac{f(tv, tk)-f(0, 0)}{t} \end{align} where $\|(v, k)\|=1$ and $v\neq 0$. Explicitly we have \begin{align} \lim_{t\rightarrow 0} \frac{1}{t}\sin\left( \frac{t^2k^2}{tv}\right)\sqrt{(tv)^2+(tk)^2}. \end{align}
