Compute the derivative: \[ f'(x) = \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5}. \] Simplify: \[ f'(x) = \frac{6}{5}(x^3 - 2x^2 - 5x + 6). \] Solve \( f'(x) = 0 \) to find critical points: \[ x^3 - 2x^2 - 5x + 6 = 0. \] Test intervals \( (-\infty, -2) \), \( (-2,1) \), \( (1,3) \), \( (3,\infty) \) by substituting values in \( f'(x) \):
Thus, \( f(x) \) is increasing on \( (-\infty, -2) \cup (1,3) \) and decreasing on \( (-2,1) \cup (3,\infty) \).
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.