Direct optimization with a Lagrange multiplier. First find the gradient exactly as before: with $u=2x+y$, $f_x=2g'(u)$, $f_y=g'(u)$, so at $(1,1)$, $u=3$ and $g'(3)=3$ gives \[\nabla f(1,1)=(6,3).\]
For a unit vector $\hat u=(a,b)$ with $a^2+b^2=1$, the directional derivative is \[D_{\hat u}f=6a+3b.\] We maximize $6a+3b$ subject to the constraint $a^2+b^2=1$. Using a Lagrange multiplier $\lambda$: \[\nabla(6a+3b)=\lambda\nabla(a^2+b^2)\ \Rightarrow\ 6=2\lambda a,\ 3=2\lambda b.\] So $a=\dfrac{3}{\lambda}$, $b=\dfrac{3}{2\lambda}$. Substituting into $a^2+b^2=1$: \[\frac{9}{\lambda^2}+\frac{9}{4\lambda^2}=1\ \Rightarrow\ \frac{45}{4\lambda^2}=1\ \Rightarrow\ \lambda^2=\frac{45}{4}\ \Rightarrow\ \lambda=\pm\frac{3\sqrt5}{2}.\] Taking $6a+3b$ to be a maximum (not a minimum) requires $\lambda>0$, so $\lambda=\dfrac{3\sqrt5}{2}$, giving \[a=\frac{3}{\lambda}=\frac{3}{3\sqrt5/2}=\frac{2}{\sqrt5},\qquad b=\frac{3}{2\lambda}=\frac{3}{3\sqrt5}=\frac{1}{\sqrt5}.\] So $\hat u=\left(\dfrac{2}{\sqrt5},\dfrac{1}{\sqrt5}\right)=\dfrac{1}{\sqrt5}(2\hat i+\hat j)$, the same result as normalizing the gradient directly. \[\boxed{\hat u=\frac{1}{\sqrt5}(2\hat i+\hat j)}\]