I tried to do the following:
$$w=\frac{1}{z}=\frac{x-iy}{x^2+y^2}$$
$\implies u = \frac{x}{x^2+y^2} and\space v = \frac{-y}{x^2+y^2}$
$\color{green}{need\space to\space transform\space the\space line\space y = 1\space into\space the\space w-plane}$
-I am not too sure about how to go about it, do i just assume that if we are dealing with $y$ then $x$ must be zero and so:
$\implies u = \frac{0}{0+1}=0 \space and\space v = -1$
-Thanks.
EDIT:
my notes says that lines should transform to circles, and circles should transform to lines, given what i have done, i have no idea how that can be possible.
We are not just dealing with $y$, and you don't only consider $x=0$. That would only tell you where one point goes, namely $i$. The equation $y=1$ describes an entire line, the horizontal line where the imaginary part is $1$. It includes the point $52+i$, for example. The problem is to determine the image under the inversion map of the set of all points of the form $x+i$.
The line $y=1$ is symmetric about the $y$-axis and $z\mapsto \frac1z$ preserves this symmetry, so $\frac1i=-i$ will lie on a vertical axis of symmetry for the image and will also have the greatest modulus of the points in the image (because $|1/z|=1/|z|$). As $x\to\infty$ along the line $y=1$, $\frac1z\to0$, so the image will approach the point $0$. The image will be in the lower half plane ($v<0$). Even if you don't have yet a proof for why the image should be a circle, if you have heard of this fact, then the above gives you enough to guess what the image is, namely a circle with a diameter from $-i$ to $0$, but without $0$ included.
That circle would have equation $u^2+(v+\frac12)^2 =\frac14$, so let's check. You have from your formulas for $u$ and $v$ that $u=\frac{x}{x^2+1}$ and $v=\frac{-1}{x^2+1}$. Therefore $$\begin{align*}u^2+\left(v+\frac12\right)^2&=\frac{x^2}{(x^2+1)^2}+\frac{(\frac12(x^2-1))^2}{(x^2+1)^2}\\ &=\frac{4x^2}{4(x^2+1)^2}+\frac{(x^2-1)^2}{4(x^2+1)^2}\\ &=\frac{4x^2+x^4-2x^2+1}{4(x^2+1)^2}\\ &=\frac{(x^2+1)^2}{4(x^2+1)^2}\\ &=\frac14. \end{align*}$$
The term "region" is usually reserved for open subsets of $\mathbb C$, not for curves like circles and lines. The line is mapped to the circle, but the half-plane regions on either side of the line get mapped to the regions inside and outside the circle.