Interior etc. of $M = \left \{ \ \left(\begin{array}{c} x \\ y \end{array}\right) \in \mathbb{R^2}: x \geq y^2 \text{ and } y > 2x-3 \right \}$ How can one find out the interior $\mathring{M}$, the closure $\overline{M}$ and the boundary $\partial M$ of this set?

$$M = \left \{ \ \left(\begin{array}{c} x \\ y \end{array}\right) \in \mathbb{R^2}: x \geq y^2 \text{ and } y &gt; 2x-3 \right \}$$

I looked at the definitions of the interior, closure and boundary, but wasn't able to find an "algorithm" as to how one can figure them out.
Can someone show me how it's done?

This is what I get when I put the functions on Desmos:
It's quite clear from your picture: the area is (topologically) just a "ball" in the plane with part of its boundary already included. The intersection area (with strict inequalities) is the interior, but the points on the edge are also topologically speaking the boundary: balls around them stick out in both parts: inside and outside. 

So on inspection: the interior is

$$\{(x,y) \in \Bbb R^2: x&gt; y^2 \text{ and } y &gt; 2x-3\}$$

the closure includes the points that are on the edges (so have $=$ in the the inequalities):

$$\{(x,y) \in \Bbb R^2: x\ge y^2 \text{ and } y \ge 2x-3\}$$

while the boundary is a union of three curves/lines (the points of intersection of $x=y^2$ and $y=2x-3$ are easily found to be $(\frac94, \frac32)$ and $(1,-1)$):

$$\{(x,y) \in \Bbb R^2: y^2=x, 0\le x \le \frac{9}{4}\} \cup \{((x,y) \in \Bbb R^2: 0\le x \le 1, y=-\sqrt{x}\}\cup \{(x,y) \in \Bbb R^2: y=2x-3, 1\le x \le \frac94\}$$
HINT. Consider the functions  $f(x,y) = x-y^2$ and $g(x,y)=y-2x.$ Then $M=f^{-1}([0, \infty)) \cap g^{-1}((-3, \infty)).$ The rest should be really easy.