Is the set $$S=\{(x,y,z)\in \mathbb{R}^3 : \frac{\exp(x+y^2-z)-1}{x^2+y^2-z^3} > 7\}$$
open, closed, or neither (in $\mathbb{R}^3$ with the Euclidian metric)?
I think I've managed to show that it is not closed by first setting $y=z=0$ and showing that there is a solution $x_* \in (0,1)$ to $e^x-1=7x^2$ by using the intermediate value theorem. Then I said any sequence $(x_n)$ converging to $x_*$ that consists of elements of $(0,x_*)$ has the property that $(x_n, 0, 0) \in S$ but $\lim_{n \to \infty}(x_n,0,0) \not\in S$ therefore $S$ is not closed.
I'm not sure if there is a better way to do this or if $S$ is open or not.
Let $W=\{(x,y,z)\in\mathbb{R}^3\,|\,z^3\neq x^2+y^2\}$. Then $W$ is an open set and if we define$$\begin{array}{rccc}f\colon&W&\longrightarrow&\mathbb R\\&(x,y,z)&\mapsto&\dfrac{\exp(x+y^2-z)-1}{x^2+y^2-z^3},\end{array}$$then, since $f$ is continuous, $f^{-1}\bigl((7,\infty)\bigr)$ is an open subset of $W$ and therefore an oepn subset of $\mathbb{R}^3$. But $S=f^{-1}\bigl((7,\infty)\bigr)$.
It is easy to prove that $S\neq\emptyset$ and that $S\neq\mathbb{R}^3$. So, since $\mathbb{R}^3$ is connected, it follows from the fact that $S$ is open that it cannot be closed.