If $f^1\left(0\right)=3, then \underset{x\to 0}{\lim }\frac{x^2}{f\left(x^2\right)-6.f\left(4x^2\right)+5.f\left(7x^2\right)}=$

$$\begin{gathered} {\mathrm{f}}^1\left(0\right)=3 \\ \underset{\mathrm{x}\to 0}{\lim } \frac{{\mathrm{x}}^2}{\mathrm{f}\left({\mathrm{x}}^2\right)-6.\mathrm{f}\left(4{\mathrm{x}}^2\right)+5.\mathrm{f}\left(7{\mathrm{x}}^2\right)} \left(\frac{0}{0} \mathrm{form}\right) \\ =\underset{\mathrm{x}\to 0}{\lim } \frac{2\mathrm{x}}{{\mathrm{f}}^1\left({\mathrm{x}}^2\right)2\mathrm{x}-6{\mathrm{f}}^1\left(4{\mathrm{x}}^2\right)8\mathrm{x}+5{\mathrm{f}}^1\left(7{\mathrm{x}}^2\right)14\mathrm{x}} \\ =\underset{\mathrm{x}\to 0}{\lim } \frac{1}{{\mathrm{f}}^1\left({\mathrm{x}}^2\right)-24{\mathrm{f}}^1\left(4{\mathrm{x}}^2\right)+35{\mathrm{f}}^1\left(7{\mathrm{x}}^2\right)} \\ =\frac{1}{{\mathrm{f}}^1\left(0\right)(1-24+35)} \\ =\frac{1}{3\left(12\right)}=\frac{1}{36} \end{gathered}$$