If $f(x,y)=x-y^{3}sin(\frac{1}{y})$ if $y\ne0$ and $f(x,y)=x$ if $y=0$.I have calculated $f_x(0,0)=1$ and $f_y(0,0)=0$.Can using this two things Am I able to comment anything about continuity or differentiability of $f(x,y)$ at $(0,0)?$
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If $f(x,y)=x-y^{3}\sin(\frac{1}{y})$ if $y\ne0$ and $f(x,y)=x$ if $y=0$.
Your calculation of partial derivatives is correct: They both exist and $f_x(0,0)=1$ and $f_y(0,0)=0$. Existence of partial derivatives does not give any information about continuity. Thus we have to use the definition of continuity at $(0, 0)$: $$\lim_{(x, y)\to (0, 0)}f(x, y)=f(0, 0).$$ This is true since $$\lim_{(x, y)\to (0, 0)}\bigg(x-y^{3}\sin(\frac{1}{y})\bigg)= f(0, 0)=0$$ by the sandwich rule. We have seen that $f$ is continuous and have partial derivatives at $(0, 0)$. For differentiability we could check the continuity of partial derivatives at the origin. Instead, we can check the following limit: $$\lim_{(h, k)\to (0, 0)}\frac{|f(h, k)-f(0, 0)-f_x(0, 0)h-f_y(0, 0)k|}{\sqrt{h^2+k^2}}=\lim_{(h, k)\to (0, 0)}\frac{|h-k^{3}\sin(\frac{1}{k})-h|}{\sqrt{h^2+k^2}}=\lim_{(h, k)\to (0, 0)}\frac{|k^{3}\sin(\frac{1}{k})|}{\sqrt{h^2+k^2}}=0$$ by the sandwich rule. Therefore, $f$ is differentiable at $(0, 0)$
The Phenotype
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daulomb
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Hints:
If the partial derivatives of $f$ exist in an open set containing $(0,0)$ and are continuous at $(0,0)$, then $f$ is differentiable at $(0,0)$.
If $f$ is differentiable at $(0,0)$, then $f$ is continuous at $(0,0)$.
The Phenotype
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it is not correct, existence of partial derivatives does not imply differentiability! – daulomb Dec 31 '17 at 21:37
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So,here $f_x=1$,then clearly it is continuous at $(0,0)$ and $f_y=ycos(\frac{1}{y})-sin(\frac{1}{y})(3y^{2})$,then I think $f_y$is also continuous – ogirkar Dec 31 '17 at 21:38
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@omkar Girkar: is this comment for me? – daulomb Dec 31 '17 at 21:39
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means $f$ is differentiable? – ogirkar Dec 31 '17 at 21:40
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@daulomb..no it was for answer. – ogirkar Dec 31 '17 at 21:40
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@omkarGirkar: if you edit your question I can say something I dont see what is f exactly – daulomb Dec 31 '17 at 21:42
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@daulomb Please read my answer again: I state that the partial derivatives must also be continuous. – The Phenotype Dec 31 '17 at 21:44
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@omkar_Girkar Correct. Of course with $f_y(x,y)=0$ for $y=0$. – The Phenotype Dec 31 '17 at 21:53