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(B) Given the curve equation: $y^{5}-9xy+2x=0$. Differentiating with respect to $x$: $5y^{4}\frac{dy}{dx} - 9(y + x\frac{dy}{dx}) + 2 = 0$. Rearrangin

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$2$

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(B) Given the curve equation: $y^{5}-9xy+2x=0$. Differentiating with respect to $x$: $5y^{4}\frac{dy}{dx} - 9(y + x\frac{dy}{dx}) + 2 = 0$. Rearranging for $\frac{dy}{dx}$: $\frac{dy}{dx}(5y^{4} - 9x) = 9y - 2$. $\frac{dy}{dx} = \frac{9y - 2}{5y^{4} - 9x}$. For the tangent to be parallel to the $x$-axis,$\frac{dy}{dx} = 0$,which implies $9y - 2 = 0$,so $y = \frac{2}{9}$. Substituting $y = \frac{2}{9}$ into the original equation: $(\frac{2}{9})^{5} - 9x(\frac{2}{9}) + 2x = 0 \Rightarrow (\frac{2}{9})^{5} - 2x + 2x = 0 \Rightarrow (\frac{2}{9})^{5} = 0$,which is impossible. Thus,there are no points where the tangent is parallel to the $x$-axis,so $M = 0$. For the tangent to be parallel to the $y$-axis,the denominator $5y^{4} - 9x = 0$,so $x = \frac{5y^{4}}{9}$. Substituting this into the original equation: $y^{5} - 9y(\frac{5y^{4}}{9}) + 2(\frac{5y^{4}}{9}) = 0 \Rightarrow y^{5} - 5y^{5} + \frac{10y^{4}}{9} = 0 \Rightarrow -4y^{5} + \frac{10y^{4}}{9} = 0 \Rightarrow y^{4}(-4y + \frac{10}{9}) = 0$. This gives $y = 0$ or $y = \frac{10}{36} = \frac{5}{18}$. If $y = 0$,then $x = 0$. If $y = \frac{5}{18}$,then $x = \frac{5}{9}(\frac{5}{18})^{4}$. Thus,there are $2$ points where the tangent is parallel to the $y$-axis,so $N = 2$. Therefore,$M + N = 0 + 2 = 2$.

Let $M$ and $N$ be the number of points on the curve $y^{5}-9xy+2x=0$,where the tangents to the curve are parallel to the $x$-axis and $y$-axis,respectively. Then the value of $M + N$ equals $......$

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