If $x = 4t^3 + 3, y = 3t^4 + 4$ and $\dfrac{\frac{d^2x}{dy^2}}{\left(\frac{dx}{dy}\right)^n}$ is constant then the value of $n$ is

Given $x = 4t^3 + 3$ and $y = 3t^4 + 4$.

Differentiating with respect to $t$:

$\dfrac{dx}{dt} = 12t^2$

$\dfrac{dy}{dt} = 12t^3$

$\dfrac{dx}{dy} = \dfrac{\dfrac{dx}{dt}}{\dfrac{dy}{dt}} = \dfrac{12t^2}{12t^3} = \dfrac{1}{t} = t^{-1}$

Now, finding the second derivative $\dfrac{d^2x}{dy^2}$:

$\dfrac{d^2x}{dy^2} = \dfrac{d}{dy}\left(\dfrac{dx}{dy}\right) = \dfrac{d}{dt}\left(t^{-1}\right) \cdot \dfrac{dt}{dy}$

$\dfrac{d^2x}{dy^2} = -t^{-2} \cdot \dfrac{1}{12t^3} = -\dfrac{1}{12}t^{-5}$

Substituting these into the given expression:

$\dfrac{\dfrac{d^2x}{dy^2}}{\left(\dfrac{dx}{dy}\right)^n} = \dfrac{-\dfrac{1}{12}t^{-5}}{(t^{-1})^n} = -\dfrac{1}{12}t^{n-5}$

For this expression to be a constant, it must be independent of $t$. Therefore, the exponent of $t$ must be zero:

$n - 5 = 0 \Rightarrow n = 5$

Answer: $5$