I want to contruct a continuous function of multivariable such that the directional derivative exists nowhere. Let $W(x)$ be the Weierstrass function, which is continuous everywhere but differentiable nowhere. We can also prove that the left and right derivatives of $W(x)$ exists nowhere. Let $f (x,y)=W (x)\cdot\sin y$, then the function $f(x,y)$ is continuous and does has directional derivative in all direction $(\cos\theta, \sin\theta)\neq(0, \pm1)$. In fact, let $$\rho=\sqrt{(\Delta x)^2+(\Delta y)^2},\quad \Delta x=\rho\cdot\cos\theta,\quad \Delta y=\rho\cdot\sin\theta,$$ then the limit \begin{eqnarray*} % \nonumber to remove numbering (before each equation) & & \lim_{\rho\to0}\frac{W(x+\Delta x)\cdot\sin(y+\Delta y)-W(x)\cdot\sin y}{\rho} \\ &=& \lim_{\rho\to0}\frac{(W(x+\Delta x)-W(x))\cdot\sin(y+\Delta y)}{\rho}+ \lim_{\rho\to0}\frac{(\sin(y+\Delta y)-\sin (y))\cdot W(x)}{\rho} \\ &=& \sin y\cdot\cos\theta\cdot\fbox{$\lim\limits_{\Delta x\to0}\dfrac{W(x+\Delta x)-W(x)}{\Delta x}$}+W(x)\cdot\cos y\cdot\sin\theta \end{eqnarray*} does not exist. However, the directional derivative on the direction $(0, \pm1)$ exists.
I guess the function $f(x,y)=W(x)+W (y)$ or $W(x)\cdot W(y)$ is continuous everywhere, and the directional derivative does not exist everywhere. In fact, let $$f(x,y)=W(x)+W (y),$$ then for $(\cos\theta,\sin\theta)\neq(\pm1,0),(0,\pm1)$(for the directions $(\pm1,0)$ and $(0,\pm1)$, the result is obvious), we have that \begin{eqnarray*} % \nonumber to remove numbering (before each equation) & & \lim_{\rho\to0}\frac{W(x+\Delta x)+W(y+\Delta y)-W(x)-W(y)}{\rho} \\ &=& \lim_{\rho\to0}\frac{W(x+\Delta x)-W(x)}{\Delta x}\cdot\frac{\Delta x}{\rho}+ \lim_{\rho\to0}\frac{W(y+\Delta y)-W(y)}{\Delta y}\cdot\frac{\Delta y}{\rho} \\ &=& \lim_{\rho\to0}\frac{W(x+\Delta x)-W(x)}{\Delta x}\cdot\cos\theta+ \lim_{\rho\to0}\frac{W(y+\Delta y)-W(y)}{\Delta y}\cdot\sin\theta \end{eqnarray*}
We know that both the limit $\lim\limits_{\rho\to0}\dfrac{W(x+\Delta x)-W(x)}{\Delta x}$ and $\lim\limits_{\rho\to0}\dfrac{W(y+\Delta y)-W(y)}{\Delta y}$ do not exist, but how to prove that the plus of the two limit does not exist?
You simply cannot infere that the directional derivative does not exist in any direction by simply knowing that the partial derivatives in the $x$ and $y$ directions do not exist. For instance, if you take $$ g(x,y)=W(x-y), $$ then the derivatives in the directions $(1,0),(0,1)$ do not exist, but the derivative in the direction $(1/\sqrt 2,1/\sqrt 2)$ exists and it is zero (as $g$ is constant in that direction).
Moreover, note that if $W$ was an odd function, then by a similar approach the function $f$ you wrote would have a vanishing directional derivative in the direction $(1,-1)$ at the origin. Clearly $W$ is not an odd function, but what I mean is that in order to prove what you want you really need to use the definition of the function $W$, not simply saying that $W$ is not differentiable anywhere (at least this is what it looks like).
I don’t know if the example you wrote has the property you want, or in general how to construct an explicit function in many variables with non existing partial derivatives.
Edit: I could give you a hint on how I would prove it. As I said, you cannot simply start from the fact that $W$ is not differentiable at any point. What would help is a property like the following one: there exists $0<\alpha<1$ such that $$ |W(x)-W(y)|\leq C|x-y|^\alpha, $$ and for every $x$ there exists constant $c$ and a sequence $y_n\to x$ such that $$ |W(x)-W(y_n)|>c|x-y_n|^\alpha. $$ With the first formula you ask for $W$ to be $\alpha$-Hölder continuous (this is true for the Weierstrass functions), while the second formula essentially says the condition is sharp (it cannot be improved); I don’t know if the second condition holds for $W$, but maybe that is known.
With the hypotheses above, consider a direction $(\cos(\theta),\sin(\theta))$ (assume $0<\theta<\pi/4$ for simplicity). Then, you simply want to prove that the function $$ g(x)=W(x)+W(\tan(\theta)x) $$ is not differentiable at a given point $x$. You can see that $$ |g(y_n)-g(x)|\geq|W(y_n)-W(x)|-|W(\tan(\theta)y_n)-W(\tan(\theta)x)|\geq $$ $$ \geq (c -C\tan(\theta)^\alpha)|x-y_n|^\alpha, $$ so that at least for $\theta$ small enough, one has that for some constant $\epsilon$ $$ |g(y_n)-g(x)|\geq \epsilon |x-y_n|^\alpha, $$ which implies that $g$ is not differentiable at $x$.
Edit 2: thinking about it, I can say for sure that you first example does not satisfy the property you are looking for. The Weierstrass function $W$ is odd around $x=1/2$, that is, $W(x)=-W(1-x)$. This implies that in your first example, $f(x,1-x)=0$ for all $x$, so that the dervative in the direction $(1/\sqrt 2,-1/\sqrt 2)$ at the point, say, $(1/2,1/2)$ exists and it is zero. Maybe this does not happen for $f(x,y)=W(x)\cdot W(y)$, I don’t know… I’ll think about it.