Is $\frac {x^4 +2y^2}{(x^2 +y^2)^{\lambda}}$ totally differentiable? Let $\lambda$ be a positive real number.
 When is the function
$$f(x,y) ＝ \begin{cases}0 &amp; \text{if }(x, y) = (0, 0) \\
\dfrac{x^4 +2y^2}{(x^2 +y^2)^{\lambda}} &amp;\text{if } (x, y) \neq (0, 0)\end{cases}$$
totally differentiable?

Find all $\lambda$ where f is totally differentiable.

This is a question from exam of Tokyo university.

https://www.ms.u-tokyo.ac.jp/kyoumu/a20170524.pdf
question No3（2）

I think there is no easy necessary and sufficient condition in this problem, so I tired to find some necessary conditions （and check that is sufficient）but I couldn't find it .

Thank you for your kind help.
Take limit with different cases

$$y= 0$$$$
\lim_{x\rightarrow 0} \frac{x^4}{x^{2\lambda}}$$

This limit is equal to 0 when $\lambda&lt;2$

Take another case $ x=0$

$$\lim_{y\rightarrow 0} \frac{2y^2}{y^{2\lambda}}$$

Now here the limit goes to zero when $$\lambda&lt;1$$

So we can summarise that when $$\lambda &lt;1$$.We took $\lambda&lt;1$ because in this range both cases limit to 0 but after $\lambda&gt;1$ limit x goes to zero but y doesn't
Take limit with different cases

$$y= 0$$$$
\lim_{x\rightarrow 0} \frac{x^4}{x^{2\lambda}}$$

This limit is equal to 0 when $\lambda&lt;2$

Take another case $ x=0$

$$\lim_{y\rightarrow 0} \frac{2y^2}{y^{2\lambda}}$$

Now here the limit goes to zero when $$\lambda&lt;1$$

So we can summarise that when $$\lambda &lt;1$$.We took $\lambda&lt;1$ because in this range both cases limit to 0 but after $\lambda&gt;1$ limit x goes to zero but y doesn't