Epsilon-Delta proof of multivariable function I have the following multivariable function: $f(x,y) = xy^2$, and I must prove it is continuous at (1,1). I have come to: $|((x-1)^2+(y-1)^2)^\frac{3}{2}-1| \le ((x-1)^2+(y-1)^2)^\frac{3}{2}+1$ (Triangle inequality) $<\delta^3+1=\epsilon$.
What can I do with this 1? Any tips are welcome and thank you in advance!
 A: I think you're a bit confused about what you're asked to do. You have to prove that given any $\epsilon > 0$, there is a $\delta > 0$ such that $d((x,y),(1,1)) < \delta \implies |xy^2 - 1| < \epsilon$. The distance function I have in mind on $\mathbb{R}^2$ is different from the one you've written down, namely $ d((x,y),(1,1)) = \sqrt{(x-1)^2 + (y-1)^2}$.
Now in general, if I write down some ugly formula and ask you to prove it's continuous, it's going to be incredibly tedious if not outright impossible to do this by divine inspiration. This is why we have theorems such as "the product and sum of continuous functions are continuous." So one way to prove this is to observe that $f_1(x,y) = x$ and $f_2(x,y) = y$ are both separately continuous functions, which is really just the same statement twice, and then write $f(x,y) = x \cdot y \cdot y$, and use this theorem.
You can find a discussion of this theorem here.
If your instruction is to use the definition and only the definition, my advice is to look long and hard at the proof of this fact, and then just plug in all the values you have in this special case. It shouldn't be that hard since all three of the factors are just projection functions.
A: One can pose this problem by considering the graph of the function as follow.
From the definition of continuity the $\varepsilon$-neighbourhood of the image $f(1,1)=1$ determines a couple of planes $z=1+\varepsilon$ and $z=1-\varepsilon$
(those planes in light yellow in the picture bellow).
The intersections of the planes $1\pm\varepsilon$ with the graph of $f$ determine two hyperbolas
$$
\left(\begin{array}{c}x\\y\\1\pm\varepsilon\end{array}\right),$$
with $xy^2=1\pm\varepsilon$.
Also a little circle of radius $\delta$, which is the border of a
$\delta$-neighbourhood centered at the preimage $(1,1)^{\top}$, when is transformed gives a closed curve $C$ viewed in the graph of $f$.
Then there is a point in the curve $C$ (the red fat point in the picture) which is the farest away from $p=(1,1,1)^{\top}$, which will be determined by considering that for the curve
$C$ can be parametrized as
$C(y)=\left(\dfrac{1+\varepsilon}{y^2}, y,1+\varepsilon\right)^{\top}$ and
the distance function
$$d(p,C(y))=\sqrt{\left(\dfrac{1+\varepsilon}{y^2}-1\right)^2+(y-1)^2+\varepsilon^2}.$$
Once you have the coordinates of this red point you can determine the coordinates of the little red point bellow in the $\delta$-circle centered a $(1,1)^{\top}$, which allows you see which is this $\delta$ conditioned by $\varepsilon$.

