$\dot{x}=-2x-2y$
$\dot{y}=-x-3y$
Equilibrium point is $(0,0)$. Eigenvalues are $\lambda_+=-1$ and $\lambda_-=-4$ which have corresponding eigenvectors $2\choose -1$ and $1 \choose 1$ respectively. The point is a stable node.
So when sketching the diagram, there will be a line through the origin with gradient $1$ and a line through the origin with gradient $-1/2$. The arrows on these lines are going to be inwards towards the equilibrium point. How do I sketch the phase paths and isoclines?
To begin with your analysis, you should first find the regions of the plane that have a zero partial derivative. Namely $\dot x, \dot y = 0$.
Setting $\dot x = -2x - 2y = 0 \implies y = -x$ is a set of points where $\dot x = 0$.
Setting $\dot y =-x - 3y = 0\ \ \implies y = -\frac{x}3$ is a set of points where $\dot y = 0$.
While a point on the line $y=-x \implies \dot x = 0$, we can evaluate $\dot y$ along this point
to get $\dot y = -x-3(-x) = 2x$, so along this line, $x>0\implies\dot y >0, x<0\implies \dot y<0.$
A similar argument can be made about evaluating $\dot x$ along the curve which has $\dot y = 0$.
Drawing these lines, and a vector field along these 'nullclines' can give a lot of information about the behavior of systems of differential equations.