Geometric Brownian Motion and Stochastic Calculus

I am trying to calculate $$d(X_t^2)$$ for Geometric Brownian Motion. I know that for GBM we have $$dX_t=\mu X_tdt + \sigma X_tdW_t$$, where $$W_t$$ is a Wiener process.

I am trying to work towards the solution which is: $$2X_tdX_t+(dX_t)^2 = 2\mu X_t^2dt + 2\sigma X_t^2dW_t + \sigma^2 X_t^2dt$$, but I am unsure how to get there.

I know that when we apply stochastic calculus to GBM we have: $$dS_t=S_t[(\mu+\sigma^2/2)dt +\sigma dW_t]$$, but when I use this in combination with what we have above and using Ito's Lemma, I'm not getting the required answer.

Any help would be great!

Recall that Ito's lemma tells you that if $$f$$ is a $$C^2$$ function then $$df(X_t) = f'(X_t) dX_t + \frac12 f''(X_t) d \langle X \rangle_t$$ where $$d\langle X \rangle_t$$ is the object you denote with the formal notation $$(dX_t)^2$$ ($$\langle X \rangle_t$$ is the quadratic variation of $$X$$). Applying this with $$f(x) = x^2$$, you get that $$dX_t^2 = 2X_t dX_t + d \langle X \rangle_t$$ Now $$dX_t = \mu X_t dt + \sigma X_t dW_t$$ and from this we also get that $$d \langle X \rangle_t = \sigma^2 X_t^2 dt$$. Substituting these in to the equation above gives the solution.