How can I solve this using the Lagrange method? 
This is what I keep doing but the answer seems to be wrong every time.

 A: What you should use for $ \ g(x) \ $ in the method is just $ \ x^2 \ + \ y^2 \ - \ 13 \ . $  Your gradient equations are correct.  You conclude from them that
$$ \lambda \ = \ \frac{3}{2x} \ = \ - \frac{2}{2y} \ \ \Rightarrow \ \ y \ = \ - \frac{2}{3}x \ \ . $$
Insert this into the constraint equation $ \  \ x^2 \ + \ y^2 \ = \ 13  $  to find values  $ \ x \ $ and $  \ y \ . $  Note that since this is equivalent to finding the intersections of a straight line with a circle, there will be two ordered-pair solutions.  Compare the values of $ \ f(x,y) \ $ at these two points to find the maximum and minimum values of that function.
[Incidentally, all the numbers in the problem come out nicely...  I might also mention that in problems with a linear function, subject to a constraint equation representing a curve with four-fold symmetry, the maxima and minima are at points on a line through the center of the curve (in this case, the origin).]
A: You have a small mistake: From $3 = \frac{6 \, \lambda}{2\,\lambda}$ you infer $3 = 3 \, \lambda$ instead of $3 = 3$.
