Let $$y = y(x)$$ be the solution of the differential equation $$(x + 1)y' - y = e^{3x}(x + 1)^2$$, with $$y(0) = \frac{1}{3}$$. Then, the point $$x = -\frac{4}{3}$$ for the curve $$y = y(x)$$ is

We consider the differential equation $$(x+1)y' - y = e^{3x}(x+1)^2$$ with $$y(0) = \frac{1}{3}$$. Rewriting it in standard form gives:
$$ y' - \frac{y}{x+1} = e^{3x}(x+1) $$ and the integrating factor is found as
$$ \text{IF} = e^{-\int \frac{1}{x+1}dx} = e^{-\ln|x+1|} = \frac{1}{x+1} $$. Multiplying through by this factor leads to
$$ \frac{d}{dx}\left[\frac{y}{x+1}\right] = e^{3x} $$ and integrating both sides yields
$$ \frac{y}{x+1} = \frac{e^{3x}}{3} + C $$. Applying the initial condition $$y(0) = \frac{1}{3}$$ gives
$$ \frac{1/3}{1} = \frac{1}{3} + C \implies C = 0 $$, so that $$y = \frac{(x+1)e^{3x}}{3}$$.

Next, to locate critical points we compute
$$ y' = \frac{e^{3x} + 3(x+1)e^{3x}}{3} = \frac{e^{3x}(3x + 4)}{3} $$. Setting $$y' = 0$$ forces $$3x + 4 = 0$$, hence $$x = -\frac{4}{3}$$. To determine its nature, we find the second derivative:
$$ y'' = \frac{3e^{3x}(3x+4) + 3e^{3x}}{3} = e^{3x}(3x + 5) $$ and evaluating at $$x = -\frac{4}{3}$$ gives
$$ y''\left(-\frac{4}{3}\right) = e^{-4}\left(-4 + 5\right) = e^{-4} > 0 $$. Since $$y'' > 0$$ at $$x = -\frac{4}{3}$$, the point is a local minimum.

Option B: a point of local minima.