$$dY\Big(X(t)\Big) = \bigg(\frac{d}{dt}Y + \frac{d}{dX}Y \cdot \mu(t) + \frac{d^{2}}{dX^{2}}Y \cdot \frac{1}{2}\sigma^{2}(t)\bigg) \cdot dt + \Big(\frac{d}{dX}Y \cdot \sigma(t)\Big) \cdot dW$$
Logically these two should cancel out
$$\require{cancel} \cancel{dY\Big(X(t)\Big)} = \bigg(\cancel{\frac{d}{dt}Y} + \frac{d}{dX}Y \cdot \mu(t) + \frac{d^{2}}{dX^{2}}Y \cdot \frac{1}{2}\sigma^{2}(t)\bigg) \cdot dt + \Big(\frac{d}{dX}Y \cdot \sigma(t)\Big) \cdot dW$$
But they do not. Moreseo they are treated differently -- $dY\Big(X(t)\Big)$ as a function and $\frac{d}{dt}Y \cdot dt$ as a constant, which makes no sense to me.
Do we assume $dX_t = X_t(\mu dt + \sigma dW_t)$ or some such? At any rate, Ito's lemma implies $$ dY(X_t) = Y'(X_t)dX_t + \frac{1}{2}Y''(X_t)d\langle X \rangle _t $$
You would cancel out $dY(X_t)$ and $Y'dt$ - they're not the same thing (one is a stochastic differential, the other is not).
For this to make sense, $Y$ is just some (at least) twice continuously differentiable map $Y:\mathbb R\to\mathbb R$. It makes no sense to distinguish between $\frac{d}{dt}Y$ and $\frac{d}{dX}Y$.
If you were working with a parametrised version of Ito's lemma, we could have $Y=Y(x,t)$ and we'd have to bring in partial derivatives.