Is the graph of $xy=1$ in $\mathbb C^{2}$ connected?

The set $S=\{\,(x,y)\in\mathbb C^2:xy=1\,\}$ is even path-connected: Given $(x_1,y_1),(x_2,y_2)\in S$, we have $x_1,x_2\in\mathbb C^\times$, which is path connected. If $\gamma\colon [0,1]\to \mathbb C^\times$ is a curve with $\gamma(0)=x_1$, $\gamma(1)=x_2$, then $\tilde\gamma\colon[0,1]\to S$, $t\mapsto (\gamma(t),1/\gamma(t))$ is a path in $S$ from $(x_1,y_1)$ to $(x_2,y_2)$.

Regarding your second point: No. The graph (which I call $S$ above) is naturally a subset of $\mathbb C^2$, not of $\mathbb C$. The variables $x,y$ in the first formula $xy=1$ are understood to be complex, whereas (which is somewhat unlucky, didactically speaking) in $(x+iy,u+iv)$ they are real numbers and the $x+iy$ is what first was $x$ and $u+iv$ is what first was $y$. This is an unfortunate mixup of notation (especially, the two $x$ appearing in $x=x+iy$ are nnot the same!) A better notation in the problem statement might have beedn helpful for you to understand it.

It is easiest to show that this is path connected.

Step 1: Prove that every point in the solution set is connected to a point where $|x|=1$. Namely, if $xy=1$, then consider multiples $x/\lambda$ and $\lambda y$, but varying $\lambda$ between $1$ and $|x|$, you can continuously change $xy=1$ into a solution where $|x|=1$.

Step 2: Observe that when $|x|=1$, $|y|=\frac{1}{|x|}=1$ as well. Therefore, it is enough to prove that the solution set where $|x|=1$ is also path connected. For every $x=e^{i\theta}$ it must be that $y=e^{-i\theta}$ (which is the only solution). By varying $\theta$, you construct a path from any point $x$ on the unit circle to any other point.