[example: Axis Tests Can Miss Two-Dimensional Behaviour]
Let $f:\mathbb{R}^2\to\mathbb{R}$ satisfy $f(0,0)=0$ and, for $(x,y)\ne(0,0)$,
\begin{align*}
f(x,y)=\frac{xy}{x^2+y^2}.
\end{align*}
We compute the two coordinate-axis difference quotients at the origin. For $t\ne0$,
\begin{align*}
f(t,0)=\frac{t\cdot 0}{t^2+0^2}=0.
\end{align*}
Hence
\begin{align*}
\frac{f(t,0)-f(0,0)}{t}=\frac{0-0}{t}=0.
\end{align*}
Taking $t\to0$ gives $\partial_x f(0,0)=0$. Similarly, for $t\ne0$,
\begin{align*}
f(0,t)=\frac{0\cdot t}{0^2+t^2}=0.
\end{align*}
Therefore
\begin{align*}
\frac{f(0,t)-f(0,0)}{t}=\frac{0-0}{t}=0,
\end{align*}
so $\partial_y f(0,0)=0$.
The diagonal approach gives different behaviour. For $x\ne0$,
\begin{align*}
f(x,x)=\frac{x\cdot x}{x^2+x^2}.
\end{align*}
Since $x^2+x^2=2x^2$ and $x^2\ne0$,
\begin{align*}
f(x,x)=\frac{x^2}{2x^2}=\frac{1}{2}.
\end{align*}
Thus along the path $(x,x)\to(0,0)$ the function values stay equal to $1/2$, while $f(0,0)=0$. The coordinate-axis partial derivatives both exist and vanish, but the function is not continuous at the origin.
[/example]