The Product map of a Lie Group is a Submersion. Problem 7.1 of Lee's Introduction to Smooth Manifolds (2nd Edition) reads:

Show that for a Lie group $G$, the multiplication map $\mu:G\times G\to G$ is a submersion (Hint: Use Local Sections).

I did the following:
Fix $g, h\in G$. Then since $T_{(g, h)}(G\times G)\cong T_g G\oplus T_h G$, we have 
$$
d\mu_{(g, h)}(X, Y)= d(\mu\circ i^h)_gX+d(\mu\circ j^g)_h Y
$$
for $X\in T_gG, Y\in T_hG$, where $i^h:G\to G\times G$ is the map defined as $i^h(x)=(x, h)$ for all $x\in G$ and similarly for $y^g$.
Thus we have
$$
d\mu_{(g, h)}(X, Y)= dR_h|_gX+dL_g|_hY
$$
Since $dR_h|_g:T_gG\to T_{gh}G$ is a linear isomorphism, we see that the rank of $\mu$ is full. So we are done.

I do not see how to do it using the hint Lee has given

Can somebody please do it using the hint?
 A: First I will explain how you might come up with the proof below on your own. The goal is to show that each point $(g,h)$ in $G\times G$ is in the image of a smooth local section of $m$, and then to apply Theorem 4.26 in Lee's book to conclude that $m$ is a smooth submersion. Playing around, it is not too difficult to come up with a smooth local section of $m$ whose image contains the identity element $(e,e)$ of the direct product $G\times G$. Then use the group multiplication and the fact that the left- and right-multiplication maps $L,R$ are diffeomorphisms to transfer this section to a more arbitrary section whose image contains a general point $(g,h)$.

Consider the element $(e,e)\in G\times G$. Let $U$ be the subgroup generated by the connected component in $G$ containing $e$. By Proposition 7.14 in Lee, $U$ is an open subset of $G$. Consider the map $$\sigma\colon U\to G\times G$$ defined by $\sigma(x) = (x,e)$. It is clear that $\sigma$ is a smooth map whose image contains $(e,e)=\sigma(e)$ such that $$(m\circ \sigma)(x) = m(x,e) = x,$$
i.e. such that $m\circ \sigma = \mathrm{Id}_U$. It follows that $\sigma$ is a smooth local section of $m$ whose image contains $(e,e)$.
Now let $(g,h)\in G\times G$ be arbitrary, and consider the set $V = gUh$ in $G$. The set $V$ is open in $G$ because it is the image of the open set $U$ under the diffeomorphism $L_g\circ R_h\colon G\to G$. Let $L_{(g,h)}(x,y) = (gx,hy)$ be the diffeomorphism $G\times G\to G\times G$ given by left-multiplication by $(g,h)$. Finally, define $\tilde \sigma\colon gUh\to G\times G$ to be the composition of smooth maps
$$
gUh\xrightarrow{\displaystyle R_{h^{-1}}\circ L_{g^{-1}}}U\xrightarrow{\displaystyle\sigma}G\times G\xrightarrow{\displaystyle L_{(g,h)}}G\times G.
$$
Then it's clear that $\tilde\sigma$ is a smooth map with $\tilde\sigma(geh) = (g,h)$ and
\begin{align*}
\big(m\circ\tilde \sigma\big)(gxh) &= m\big(\tilde\sigma(gxh)\big) \\
&= m\big(L_{(g,h)}\big(\sigma\big(R_{h^{-1}}\big(L_{g^{-1}}(gxh)\big)\big)\big)\big) \\
&= m\big(L_{(g,h)}\big(\sigma(x)\big)\big) \\
&= m\big(L_{(g,h)}(x,e)\big) \\
&= m(gx,h)  = gxh,
\end{align*}
i.e. $m\circ\tilde\sigma = \mathrm{Id}_{gUh}$. Hence $\tilde \sigma$ is a smooth local section of $m$ whose image contains $(g,h)$. Since $(g,h)\in G\times G$ was arbitrary, we conclude that each point in $G\times G$ is in the image of a smooth local section of $m$. Hence $m$ is a smooth submersion by Theorem 4.26 in Lee, as desired.
