Using expansions, we get,
$\underset{\mathrm{x}\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\mathrm{a}\left(1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+..\right)+\mathrm{b}\left(1-\frac{{\mathrm{x}}^2}{2!}+...\right)+\mathrm{c}+\mathrm{d}\mathrm{x}}{\mathrm{x}{\left(\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+.....\right)}^2}=3$
$\underset{\mathrm{x}\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)+\left(\mathrm{a}+\mathrm{d}\right)\mathrm{x}+\left(\frac{\mathrm{a}-\mathrm{b}}{2}\right){\mathrm{x}}^2+\frac{\mathrm{a}}{6}{\mathrm{x}}^3+\dots }{{\mathrm{x}}^3{\left(1-\frac{{\mathrm{x}}^2}{3!}+.....\right)}^2\mathrm{}}=3$
$\because$ in the denominator lowest power of $\mathrm{x}$ is $3$
For the limit to be finite, the numerator should also have the least power of $\mathrm{x}$ as $3$
$\therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=0\hspace{0.17em}\hspace{0.17em}\mathrm{}...\left(1\right)$
$\mathrm{a}+\mathrm{d}=0\mathrm{}...\left(2\right)$
$\frac{\mathrm{a}-\mathrm{b}}{2}=0\mathrm{}...\left(3\right)$
Now, $\frac{\left(\frac{\mathrm{a}}{6}\right)}{1}=3\Rightarrow \mathrm{a}=18$
From $\left(1\right),\left(2\right),\left(3\right)$, we get,

$\mathrm{a}=18, \mathrm{b}=18, \mathrm{c}=–36, \mathrm{d}=–18$
$\frac{\mathrm{a}\mathrm{b}\mathrm{d}}{{\mathrm{c}}^3}=\frac{-{\left(18\right)}^3}{-8{\left(18\right)}^3}=\frac{1}{8}$