Given,

$y^5-9xy+2x=0 .......\left(1\right)$

Now differentiating both side we get,

$5y^4\frac{dy}{x}-9x\frac{dy}{dx}-9y+2=0$

$\Rightarrow \frac{dy}{dx}\left(5y^4-9x\right)=9y-2$

$\Rightarrow \frac{dy}{dx}=\frac{9y-2}{5y^4-9x}=0 .....\left(2\right)$ 

$\Rightarrow 9y-2=0$ (for horizontal tangent)

$\Rightarrow y=\frac{2}{9}$, which does not satisfy the original equation so $M=0$.

As ${\left(\frac{2}{9}\right)}^5-9x\times \frac{2}{9}+2x=0$

$\Rightarrow {\left(\frac{2}{9}\right)}^5-2x+2x=0$

$\Rightarrow {\left(\frac{2}{9}\right)}^5=0 \left\{\text{which is a false statement}\right\}$ 

Now $5y^4-9x=0$  (from equation $\left(2\right)$ for vertical tangent)

$\Rightarrow 5y^4\left(9y-2\right)-9y^5=0$ (putting the value of $x$ from equation $\left(1\right)$)

$\Rightarrow y^4\left[45y-10-9y\right]=0$

$\Rightarrow y=0$ or $36y=10$$\Rightarrow y=\frac{5}{18}$

So at $y=0\Rightarrow x=0$ and $y=\frac{5}{18}\Rightarrow x=x_1$

$\begin{array}{cc}\left(0,0\right)& \left(x_1,\frac{5}{18}\right)\end{array}$ (where $x_1=\frac{{\left({\displaystyle \frac{5}{18}}\right)}^5}{9\times {\displaystyle \frac{5}{18}}-2}$)

So, $N=2$

$M+N=0+2=2$