Determine if $\frac{x^2y}{x^2+y^2}$ is differentiable at $(x,y)=(0,0)$

Let $$k$$ be a real number. Define $$f\colon\Bbb{R}^2\to\Bbb{R}$$ by $$f(x,y)=\begin{cases}\dfrac{x^ky}{x^2+y^2}&\text{if (x,y)\neq(0,0)},\\k-2&\text{if (x,y)=(0,0)}.\end{cases}$$

1. Find the value of $$k$$ such that $$f$$ is continuous at $$(0,0)$$.
2. For the value of $$k$$ found in part (1), determine whether $$f$$ is differentiable at $$(0,0)$$.

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So, I have solved the part (1) and found that the $$k=2$$. Then both partial derivatives $$f_x(0,0)$$ and $$f_y(0,0)$$ of the function $$f(x,y)$$ equal $$0$$.

I know that if the partial derivatives exist and continuous at $$(0,0)$$, then the function is differentiable at $$(0,0)$$.

However, I also know that we can check the differentiability using this formula:

Definition. Let $$f\colon X\to\Bbb{R}$$ where $$X\subset\Bbb{R}^2$$ is open, and let $$\mathbf{a}\in X$$. Suppose $$f_x(\mathbf{a}),f_y(\mathbf{a})$$ exist. We say that $$f$$ is differentiable at $$\mathbf{a}$$ if $$\lim_{\mathbf{x}\to\mathbf{a}}\frac{f(\mathbf{x})-[f(\mathbf{a})+f_x(\mathbf{a})(x_1-a_1)+f_y(\mathbf{a})(x_2-a_2)]}{\|\mathbf{x}-\mathbf{a}\|}=0.$$

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And using, this formula I get that it is not continuous at $$(0,0)$$ (I have used $$x=r\cos\alpha$$ and $$y = r\sin\alpha$$ substitution to check it.)

Can you please tell me which approach is right, and is the function differentiable at $$(0,0)$$ if $$k=2$$?

• Welcome to math.SE!! – manooooh Dec 26 '19 at 21:18
• Yes, $k=2$ is correct. You mean to say that $f$ is not differentiable at $(0,0)$? Please show the details of how you showed that limit is not $0$. – Ted Shifrin Dec 26 '19 at 22:20

$$0 = \lim_{(a_1,a_2) \to (0,0)} \frac{f(a_1,a_2)}{\|(a_1, a_2)\|} = \lim_{(a_1,a_2) \to (0,0)} \frac{a_1^2a_2}{(a_1^2+a_2^2)^{3/2}}$$
If we approach $$0$$ along the line $$y = x$$ in the first quadrant, we get
$$\frac{f(t,t)}{\|(t,t)\|} = \frac{t^3}{(2t^2)^{3/2}} = \frac1{2^{3/2}}$$
which doesn't converge to $$0$$ as $$t \to 0^+$$ and therefore the above limit is not $$0$$.
We conclude that $$f$$ is not differentiable at $$(0,0)$$.