If normal to the curve y=f(x) at x=0 be given by the equation $9x-y+3=0$, then the value of $\underset{\mathrm{x}\to 0}{\lim }{\mathrm{x}}^2{\{\mathrm{f}\left({\mathrm{x}}^2\right)-6\mathrm{f}\left(4{\mathrm{x}}^2\right)+5\mathrm{f}\left(6{\mathrm{x}}^2\right)\}}^{-1}$ is

Given, $\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{f}}^{\mathrm{I}}\left(\mathrm{x}\right) \therefore {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=0}={\mathrm{f}}^{\mathrm{I}}\left(0\right)$

$\therefore \mathrm{Slope}\mathrm{of}\mathrm{normal}=-\frac{1}{{\mathrm{f}}^{\mathrm{I}}\left(0\right)}=9\Rightarrow {\mathrm{f}}^{\mathrm{I}}\left(0\right)=-\frac{1}{9}$

Now, $\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{x}}^2}{\mathrm{f}\left({\mathrm{x}}^2\right)-6\mathrm{f}\left(4{\mathrm{x}}^2\right)+5\mathrm{f}\left(6{\mathrm{x}}^2\right)}$

$\begin{array}{l}=\underset{x\to 0}{lim} \frac{x}{f\left(x\right)-6f\left(4x\right)+5f\left(6x\right)}\left[\mathrm{Re}\text{ placing }{x}^2\text{ by }x\right]\\ =\underset{x\to 0}{lim} \frac{x}{\left(f\right(x)-f(0\left)\right)-6\left(f\right(4x)-f(0\left)\right)+5\left(f\right(6x)-f(0\left)\right)}\\ =\underset{x\to 0}{lim} \frac{1}{\frac{f\left(x\right)-f\left(0\right)}{x-0}-6\frac{\left(f\right(4x)-f(0\left)\right)}{x-0}+5\frac{\left(f\right(6x)-f(0\left)\right)}{x-0}}\\ =\underset{x\to 0}{lim} \frac{1}{\frac{f\left(\right(x)-f(0\left)\right)}{x-0}-6\times 4\left(\frac{f\left(4x\right)-f\left(0\right)}{4x-0}\right)+5\times 6\left(\frac{f\left(6x\right)-f\left(0\right)}{6x-0}\right)}\\ =\frac{1}{f^I\left(0\right)-24f^1\left(0\right)+30f^I\left(0\right)}=\frac{1}{7f^I\left(0\right)}=\frac{1}{7}\times (-9)=-\frac{9}{7}\end{array}$