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Thought about it a little and couldn't find an answer. If:

$u(t)=f(t,c)$

$v(t)=f(c,t)$

are both continuous for every $c\in\mathbb{R}$, then $f : \mathbb{R^2 \to \mathbb{R}}$ is continuous too.

Tried to use the triangle inequality this way:

$|f(x,y)-f(a,b)| \leq |f(x,y)-f(x,b)+f(x,b)-f(a,y)+f(a,y)-f(a,b)| \leq$

$\leq |f(x,y)-f(x,b)| + |f(x,b)-f(a,y)| + |f(a,y)-f(a,b)|$

The left and right parts could be bounded, but I can't think about a reason why $|f(x,b)-f(a,y)|$ could be bounded either. Hence, maybe I can't claim that?

Will appreciate your help :)

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    this is not true, however if you add the extra condition that $f$ maps compact sets to compact sets then it is. – 2017-01-14
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    Do you have an example? – 2017-01-14
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    what about $f(x,y)=\frac{xy}{x^2+y^2}$ $f(0,0)=0$? – 2017-01-14
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    Classic, thanks – 2017-01-14

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