$\newcommand\R{\mathbb{R}}$ Let $f\colon \R^n \to \R^m$. $f$ is partial continuous in $x=(x_1,\ldots,x_n)\in \R^n$ iff $f_k\colon\R \to \R^m, x' \mapsto f(x_1,\ldots,x_{k-1},x',x_{k+1}\ldots,x_n)$ is continuous for all $1 \le k \le n$. My question: If $f$ is partial continuous in every $x \in \R^n$, does this imply the continuity of $f$?
Additional information: The inverse is clear to me. I can also construct an $f$ which is is partial continuous in a specific point (but not every point), but not continuous in this point (take $$f\colon \R^2\to \R,\quad f(x_1,x_2) = \begin{cases}0 & \text{if $x_1x_2=0$}\\1 & \text{otherwise}\end{cases}$$ and $x=(0,0)$). So I wonder whether there exists a counterexample for the "global" case described above.
$\newcommand\R{\mathbb{R}}$ @Chappers provided a link to a counterexample. Let $f\colon \R^2 \to \R$ by $$f(x_1,x_2) = \frac{x_1x_2}{x_1^2+x_2^2}$$ and $x = (0,0)$. $f$ is obviously partial continuous in every point (just fix $x_1$ or $x_2$). But let $(a_k)_{k\ge0}$ via $a_k = \left(\frac{\sin(k)}k,\frac{\cos(k)}k\right)$. Then $$\lim\limits_{k\to\infty} a_k = (0,0) = x,$$ but $$\lim\limits_{k\to \infty} f(a_k) = \lim\limits_{k\to \infty}\sin(k)\cos(k) = \lim\limits_{k\to \infty} \frac12\sin(2k)$$ does not exist, so $f$ is not continuous. So the answer to my question is "wrong".