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Let $M := (0,infty) subset mathbb{R^2}$ and $f:mathbb{R}^2 to mathbb{R}$.

How can one prove that all directional derivatives $frac{partial f}{partial e}$ of $f(x,y)$ are existing in the origin and calculate them?

Is the following correct?

Let $(a_n)_{ninmathbb{R}}$ be a sequence with $a_n = (frac{1}{n},frac{1}{n})$.

If I use $a_n$ in the function $f(x,y)$, I get

$$frac{sin(frac{1}{n}+frac{1}{n})^3}{(frac{1}{n})^2+(frac{1}{n})^2} $$

Then $lim n to infty = 0$

Therefore all directional derivatives exist in the origin.

I know that

$$frac{partial f}{partial x} = dfrac{3x^2cosleft(x^3+y^3right)}{x^2+y^2}-dfrac{2xsinleft(x^3+y^3right)}{left(x^2+y^2right)^2}$$

and

$$frac{partial f}{partial y} = dfrac{3y^2cosleft(y^3+x^3right)}{y^2+x^2}-dfrac{2ysinleft(y^3+x^3right)}{left(y^2+x^2right)^2}$$

By using the sequence $a_n$ we get $lim n to infty = 0$

So $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$ exist in the point $(0,0)$

Is that correct? I didn't use $frac{partial f}{partial e}$

No, the reasoning is not correct. Your discussion merely partially shows that when $boldsymbol e = (1/sqrt 2, 1/sqrt 2)$, $partial f/partial boldsymbol e$ might exist.

Just apply the definition: the directional derivative of $f$ along the direction $boldsymbol e$ at $boldsymbol x in mathbb R^2$ is
$$
lim_{tto 0} frac {f(boldsymbol x + tboldsymbol e) - f(boldsymbol x)}t.
$$
Given any unit vector $boldsymbol e =( cos theta, sin theta)$, the limit at $boldsymbol x = (0,0)$ is [WRONG EXPONENT HERE]
$$
lim_{tto 0}frac {sin(t^3(cos^3theta + sin^3theta))}{t^2 (sin^2theta + cos^2theta)} = lim_{tto 0}frac {sin(t^3(cos^3theta + sin^3theta))}{t^2}
$$
use $|sin x | leqslant |x|$ when $|x| < 1$, we have
$$
leftvert frac {sin(t^3(cos^3theta + sin^3theta))}{t^2} rightvert leqslant |t| |cos^3theta +sin^3theta| leqslant 2|t| xrightarrow{t to 0} 0,
$$
so the limit exists for all direction $boldsymbol e$ by the squeeze theorem, and $(partial f/partial boldsymbol e) (0,0) = 0$.

EDIT

The limit should be
$$
lim_{tto 0}frac {sin(t^3(cos^3theta + sin^3theta))}{t^{color{red}3} (sin^2theta + cos^2theta)} = lim_{tto 0}frac {sin(t^3(cos^3theta + sin^3theta))}{t^{color{red}{3}}},
$$
Now use $sin (x) sim x[x to 0]$, we conclude that
$$
frac {partial f}{partial boldsymbol e} = sin^3theta + cos^3theta,
$$
not $0$ as above.

It is not correct. We have $f(a_n) ne frac{sin(frac{1}{n}+frac{1}{n})^3}{(frac{1}{n})^2+(frac{1}{n})^2}$ !

With the definition: $frac{partial f}{partial e}(0,0)= lim_{t to 0}frac{f(te)-f(0,0)}{t}$, if the limit exists.

Let $e=(u,v)$ be a direction. WLOG we can assume that $u^2+v^2=1$. Then we have

$frac{f(te)-f(0,0)}{t}=frac{sin(t^3(u^3+v^3)}{t^3}$.

If $u^3+v^3=0$, then $frac{partial f}{partial e}(0,0)=0$.

Now assume that $u^3+v^3 ne 0$. It follows that

$frac{f(te)-f(0,0)}{t}=frac{sin(t^3(u^3+v^3)}{t^3}=(u^3+v^3)frac{sin(t^3(u^3+v^3)}{t^3(u^3+v^3)} to u^3+v^3$ as $t to 0$, since $frac{sin x}{x} to 1$ as $x to 0$.

Conclusion:

$frac{partial f}{partial e}(0,0)=u^3+v^3$.