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Explain implications among these assertions:

a. The function $f:\mathbb R^2\rightarrow \mathbb R$ is continuously differentiable.
b. The function $f:\mathbb R^2\rightarrow \mathbb R$ has directional derivatives in all directions at each point in $\mathbb R^2$.
c.The function $f:\mathbb R^2\rightarrow \mathbb R$ has first order partial derivatives at each point in $\mathbb R^2$.
(Hint: for $b\nRightarrow a$ consider $f=\begin {cases} sin(\frac{y^2}{x})\sqrt{x^2+y^2}, x\neq0 \\0,x=0 \end {cases}$ ).

So, if $f$ is continuously differentiable, then, obviously, it has first order partial derivatives, so $a \Rightarrow c$, however the converse is not true.
Following the hint consider $$\frac{\partial f}{\partial \vec{p}}(\vec{0})=\lim_{t\rightarrow 0}\frac{f(tp_1,tp_2)-f(0,0)}{t}=\lim_{t\rightarrow 0}\frac{sin(\frac{(tp_2)^2}{tp_1})\sqrt{t^2(p_1^2+p_2^2)}}{t}=\lim_{t\rightarrow 0}\frac{sin(\frac{tp_2^2}{p_1})|t|\sqrt{(p_1^2+p_2^2)}}{t}=0$$

Also $\frac{\partial f}{\partial x} (x,y)=\frac{x^3 sin(\frac{y^2}{x}) - y^2 (x^2 + y^2) cos(\frac{y^2}{x})}{x^2\sqrt{x^2 + y^2}}$, and $\frac{\partial f}{\partial x} (0,0)=\frac{f(t,0)-f(0,0)}{t}=0.$ But $$\lim_{(x,y)\rightarrow (0,0)}\frac{\partial f}{\partial x} (x,y)=\text{(restricted to the line } x=my^2)=\lim_{y\rightarrow 0} \frac{m^3 y^6 sin(\frac{1}{m}) - y^2 (m^2 y^4 + y^2) cos(\frac{1}{m}))}{m^2 y^4 \sqrt{m^2 y^4 + y^2}} = -\infty \neq 0.$$ So $\frac{\partial f}{\partial x} (x,y)$ is not continuous at (0,0).

Are there any mistakes in my prove? And I don't know how to prove/disprove other implications, can somebody show how to prove or give counterexamples to these implications?

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