\(f(x)=\frac{1}{3x^2}=\frac{1}{3}x^{-2}\\ f'(x) =\frac{-2}{3}x^{-3}\\ f'(-1) =\frac{-2}{3}\times (-1)^{-3}\\ f'(-1) =\frac{-2}{3*(-1)^{3}}\\ f'(-1)=\frac{-2}{3*-1}\\ f'(-1)=\frac{2}{3}\\ \)
Since the gradient at x=-1 is positive, the graph is increasing.
You can also see this from PM's graph. you can see that the gradient of the tangent to the curve is positive when x=-1