Let $f$ be a real-valued differentiable function on $\mathbb{R}$ (the set of all real numbers) such that $f(1) = 1$. If the $y$-intercept of the tangent at any point $P(x, y)$ on the curve $y = f(x)$ is equal to the cube of the abscissa of $P$, then find the value of $f(-3)$

$f \text{ differentiable on } \mathbb{R}, \qquad f(1) = 1$ $\text{Tangent at } P(x, y) \text{ on } y = f(x):\ y\text{-intercept} = x^{3}$ $\text{Build the ODE, find } f(x), \text{ then } f(-3)$ $\text{Tangent at } P(x, y):\ Y - y = f'(x)(X - x)$ $Y\text{-intercept } (X = 0):\quad b = y - x\,f'(x)$ $b = x^{3} \Rightarrow y - x\,y' = x^{3}$ $y' - \dfrac{y}{x} = -x^{2}$ $\dfrac{d}{dx}\!\left(\dfrac{y}{x}\right) = \dfrac{x\,y' - y}{x^{2}} = -x$ $\dfrac{y}{x} = -\dfrac{x^{2}}{2} + C \Rightarrow y = Cx - \dfrac{x^{3}}{2}$ $1 = C - \dfrac{1}{2} \Rightarrow C = \dfrac{3}{2}$ $\boxed{f(x) = \dfrac{3x - x^{3}}{2}}$ $f(-3) = \dfrac{3(-3) - (-3)^{3}}{2} = \dfrac{-9 + 27}{2} = 9$ $\boxed{f(-3) = 9}$ $f'(x) = \dfrac{3}{2} - \dfrac{3x^{2}}{2}$ $y - x f'(x) = \dfrac{3x - x^{3}}{2} - x\!\left(\dfrac{3}{2} - \dfrac{3x^{2}}{2}\right) = x^{3}\ \text{verified}$ $f(x) = \dfrac{3x - x^{3}}{2}; \qquad f(-3) = 9$