Is there a mistake in the problem? Continuity of a two-variable function. Let $\phi : \mathbb{R} \rightarrow \mathbb{R}$ be a $C^2$ function, such that $\phi (0) = 0$ and $\phi''(0) \neq 0$. If
$$
f(x,y) = 
\begin{cases}
 \frac{x\phi(y) - y\phi(x)}{x^2+y^2}, &\text{if } (x,y) \neq (0,0) \\
 0, &\text{if } (x,y) = (0,0)
\end{cases},
$$
then show that $f$ is continuous, but not $C^1$.
I have tried solving it. The problem is, that if $\phi(x)=\sqrt{x}$, then $f$ would not be continuous (taking the path $y=2x$, and letting $x \rightarrow 0$ makes the limit $-\infty \neq 0$). Did I not understand the question properly? Is taking $\phi(x) = \sqrt{x}$ as a counter example a mistake? Is there some mistake with the problem? And if not, could I get some sort of hint on solving it, as I have tried a few methods, but I can't seem to solve it.
Edit: I may have solved it, but I am not sure if it is correct.
Let $\epsilon > 0$. We have
\begin{align}
|f(x,y) - f(0,0)| &= \frac{|x\phi(y)-y\phi(x)|}{x^2+y^2} \\
&\leq \frac{|x||\phi(y)|+|y||\phi(x)|}{x^2+y^2} \\
&\leq \frac{\sqrt{x^2+y^2}(|\phi(y)| + |\phi(x)|)}{x^2+y^2}, \quad \text{since }|x|,|y|\leq\sqrt{x^2+y^2} \\
&= \frac{|\phi(x)| + |\phi(y)|}{\sqrt{x^2+y^2}}.
\end{align}
For any of those $x,y \in \mathbb{R}^*$, because $\phi$ is continuous at $0$ with $\phi(0) = 0$, by having $\epsilon_1 = y^2 > 0$ and $\epsilon_2 = x^2 > 0$, there must be some $\delta_1, \delta_2 > 0$, such that if $|x|\leq\delta_1$ and $|y|\leq\delta_2$, then $|\phi(x)|\leq\epsilon_1=y^2$ and $|\phi(y)|\leq\epsilon_2=x^2$. From this, we pick $\delta_m = \min\{\delta_1, \delta_2\}$ and have $|x|,|y|\leq\delta_m$.
Continuing
\begin{align}
\frac{|\phi(x)| + |\phi(y)|}{\sqrt{x^2+y^2}} &\leq \frac{x^2+y^2}{\sqrt{x^2+y^2}} \\
&= \sqrt{x^2+y^2} \\
&\leq |x| + |y| \\
&\leq 2\delta_m
\end{align}
We pick $\delta = \min\{\epsilon, 2\delta_m\}$, making $f$ continuous at $(0,0)$.
Could someone tell me if taking $\epsilon_1 = y^2$ and $\epsilon_2 = x^2$ was a mistake?
 A: You cannot let $\epsilon_1=x^2$ and then apply the epsilon-delta definition of continuity. The $\epsilon_1$ you use must be a fixed number, not function of $x$.
As soon as you applied the inequality
$$
|f(x,y) - f(0,0)|
\leq \frac{|x||\phi(y)|+|y||\phi(x)|}{x^2+y^2},
$$
you were doomed to failure. The problem is that there that $\phi'(x)$ may very well be nonzero, so that $\phi(x)$ and $\phi(y)$ are approximately linear in $x,y$ near zero, meaning the numerator and the denominator are both quadratic in $x$ and $y$. Thus, as $x,y\to0$, you approach a constant, but it may not be zero.
Before applying the upper bound, you had the numerator was $x\phi(y)-y\phi(x)$. There, the linear parts of $\phi$ cancel out, making the numerator go to zero at a faster order and allowing you to show the limit is zero.
How to make all this rigorous? By Taylor's theorem, given $\epsilon>0$, you can choose $\delta>0$ so $|x|<\delta$ implies $|\phi(x)-\phi(0)-\phi'(0)x|=|\phi(x)-\phi'(0)x|\le \epsilon |x|$. Then when $|x|+|y|<\delta$,
\begin{align}
|f(x,y) - f(0,0)| &= \frac{|x\phi(y)-y\phi(x)|}{x^2+y^2} \\
&\leq \frac{|x\phi'(0)y-y\phi'(0)x|+|x(\phi(y)-\phi'(0)y)|+|y(\phi(x)-\phi'(0)x)|}{x^2+y^2} \\
&\leq  \frac{0+\epsilon(|x||y|+|y||x|)}{x^2+y^2}.
\end{align}
and I think you can take it from here.
