Okey if $\phi(x,y)=\ln(x^2+y^2), (x,y) \neq (0,0)$. Find the field lines for $\mathbf{G}=\nabla \phi$.
So $\mathbf{G}=\frac{2x}{x^2+y^2}\mathbf{i}+\frac{2y}{x^2+y^2}\mathbf{j}$ right? To find the field lines we use:
$$\begin{align} \frac{dx}{\frac{2x}{x^2+y^2}}&=\frac{dy}{\frac{2y}{x^2+y^2}}, \bigg|\times \frac{2}{x^2+y^2} \\ \frac{dx}{x}&=\frac{dy}{y} \\ \log|Ax|&=\log|y|\\ y&=Ax\end{align} $$
So the field lines are lines thru origo? When graphing the lines how does I know the magnitude of the lines? (the graph with small arrows)
You can see it directly from the formula. $$G(x,y) = \frac{2}{|(x,y)|^2}(x,y) = \frac{2}{r^2}(x,y)$$ (where $r$ is the magnitude of $(x,y)$) so the field vector at $(x,y)$ points in the same direction as the radius vector to the point, i.e., directly away from the origin at all points $\neq (0,0)$. You can see the magnitude of the field vector at $(x,y)$ is $\frac2r$ as well.