Question: Hi I'm trying to manipulate; $\frac{\pi\cos(\pi x)}{\sqrt{\cos^2(\pi x)}}$; to; $\pi \sqrt{\cos^2(\pi x)} \sec(\pi x)$. ~ (From Wolfram but no steps).

When I try I get $\frac{\pi\cos(\pi x)}{\sqrt{\cos^2(\pi x)}} = \frac{1}{\sqrt{\cos^2(\pi x)}}\cdot \frac{\pi}{\sec(\pi x)} = \frac{\pi}{\sqrt{\cos^2(\pi x)}sec(\pi x)}$.

I'm unsure what else I can do.

~ Thanks.

Answer: 
Perhaps surprisingly, no. A standard counter-example is
$$f(x,y)=\left\{\begin{array}{ll} 0 & \text{ if } (x,y)=(0,0) \\ \dfrac{xy^2}{x^2+y^4} & \text{ if } (x,y)\not=(0,0) \end{array}\right.$$
$f$ is not continuous at $(0,0)$ since $f(y^2,y)=\dfrac{1}{2}$ for every $y\not=0$ while $f(0,0)=0$.
**Hint:** Consider $$f(x,y)=\left\{\begin{array}{ll}1,&y\geq\sqrt{|x|},\,x\neq0\\0,&\text{otherwise} \end{array}\right.$$