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I am dealing with a continuous function $ f \colon \mathbb{R}^n \to \mathbb{R}$.

For some $ x \in \mathbb{R}^n $ I find all partial derivatives to be $0$ and $ f(x + \epsilon \cdot e_i) > f(x) $, i.e. I have a local minimum in each dimension.

I know that $ grad(x) = 0 $, hence all directional derivates in $x$ are $0$ too. Can I conclude that there is a local minimum in x, or can there be some direction $a$ with $ f(x + \epsilon \cdot a) < f(x) $ ?

If so, how would I go about proving that? If not, what would a counter example look like?

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    Try $$f(x,y)=x^2+y^2+3xy$$ – 2017-01-05
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    @Did Are we talking about $x = (0,0)$? That would violate my condition of $f(x+ \epsilon \cdot e_i) > f(x)$. – 2017-01-05
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    Sorry? Note that $f(x,0)>f(0,0)$ for every $x\ne0$ and $f(0,y)>f(0,0)$ for every $y\ne0$. – 2017-01-05
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    $ f(x,y) = x^2 + y^2 + 3xy $ works, I guess. But you had $ f(x,y) = xy $ before, which is when I wrote my comment. Thanks anyway! – 2017-01-05

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Did's example answers the question, but even more strongly:

There exist polynomial functions having a strict local minimum at the origin along every line (not just along the coordinate axes), but not having a local minimum at the origin.

The standard example is $$ f(x, y) = (y - x^{2})(y - 3x^{2}). $$ In the graph, points $(x, y)$ "below sea level" (blue) satisfy $x^{2} < y < 3x^{2}$. For every line through the origin, some deleted interval about $(0, 0)$ lies entirely outside the blue region, i.e., "above sea level". On the other hand, $f(x, 2x^{2}) < 0$ for all $x \neq 0$.

A function with a minimum along every line, but no minimum

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    Thanks, that's a great example. – 2017-01-21