Showing that $f(x,y)=\dfrac{x}{y}$ is continuous when $x>0$ and $y>0$. So I am hoping to show that $f(x,y)=\dfrac{x}{y}$ is continuous when $x>0$ and $y>0$.
I am not sure how to approach this problem. My idea was that taking $$\dfrac{\partial f}{\partial x}=\dfrac{1}{y}$$ and $$\dfrac{\partial f}{\partial y}=\dfrac{-x}{y^2}$$
My logic was that since both partials exists and are defined if $x>0$ and $y>0$, but I then discovered that existence of partial derivatives does not imply continuity. 
I am curious if there are any clever ways to show continuity of this function? Note I am not concerned with the case that $x=0$ or $y=0$
 A: First show by definition that $\pi_1(x,y)=x$ and $\pi_2(x,y)=y$ are both continuous on ${\bf R}^2$ and in particular on $S:=\{(x,y)\in{\bf R}^2\mid x>0,y>0\}$. 
Now note that $f$ is the product of two continuous functions:
$$
f(x,y)=\pi_1(x,y)\cdot g(\pi_2(x,y))
$$
where $g(z):=\frac{1}{z}$ is a function of a single variable. 

[Added:] One needs to show that the composition of two continuous functions is continuous, which in particular gives the continuity of the function $g\circ\pi_2$. Also, one should be able to show that the product of two continuous functions is continuous. 
A: For $f(x)$ and $g(y)$ continuous then in $(x_0,y_0)$ where $g(y_0)\neq0$ we have:


*

*$\forall \varepsilon>0,\exists \delta_1>0\mid |x-x_0|<\delta_1\implies |f(x)-f(x_0)|<\varepsilon$

*$\forall \varepsilon>0,\exists \delta_2>0\mid |y-y_0|<\delta_2\implies |g(y)-g(y_0)|<\varepsilon$


Since $g(y_0)\neq 0$ it is possible to choose $0<\varepsilon<\frac 12 |g(y_0)|$ 
So for $\delta=\min(\delta_1,\delta_2)$ we have
$\begin{array}{ll}
\displaystyle \left|\frac{f(x)}{g(y)}-\frac{f(x_0)}{g(y_0)}\right| &=\displaystyle\left|\frac{f(x)g(y_0)-f(x_0)g(y)}{g(y)g(y_0)}\right| =\displaystyle\left|\frac{g(y_0)(f(x)-f(x_0))-f(x_0)(g(y)-g(y_0))}{(g(y)-g(y_0))g(y_0)+g(y_0)^2}\right|\\\\
&\displaystyle<\frac{\varepsilon\left(|f(x_0)|+|g(y_0)|\right)}{\bigg||g(y_0)^2|-|g(y)-g(y_0)||g(y_0)|\bigg|}<\frac{\varepsilon\left(|f(x_0)|+|g(y_0)|\right)}{\frac 12|g(y_0)^2|}<k\,\varepsilon\end{array}$
with $k$ constant, thus the quotient is continuous in $(x_0,y_0)$.
