Differentiable Function Defined by Tangent Y-Intercept Equal to Abscissa Cubed

To find the value of $f(-3)$, we first construct the differential equation representing the given condition.

Let $P(x, y)$ be any point on the curve $y = f(x)$. The equation of the tangent line to the curve at point $P(x, y)$ is given by:
$Y - y = \frac{dy}{dx}(X - x)$

To find the $y$-intercept of this tangent line, we set $X = 0$:
$Y_{\text{intercept}} - y = \frac{dy}{dx}(0 - x) \implies Y_{\text{intercept}} = y - x\frac{dy}{dx}$

According to the problem statement, the $y$-intercept is equal to the cube of the abscissa of point $P$, which is $x^3$:
$y - x\frac{dy}{dx} = x^3$

Rearranging this differential equation:
$x\frac{dy}{dx} - y = -x^3$

Dividing both sides by $x^2$ (for $x \neq 0$):
$\frac{x\frac{dy}{dx} - y}{x^2} = -x$

The left-hand side is the derivative of $\frac{y}{x}$ with respect to $x$:
$\frac{d}{dx}\left(\frac{y}{x}\right) = -x$

Integrating both sides with respect to $x$:
$\frac{y}{x} = \int -x \, dx$
$\frac{y}{x} = -\frac{x^2}{2} + C$
$y = f(x) = -\frac{x^3}{2} + Cx$

We are given the initial condition $f(1) = 1$. Substituting $x = 1$ and $y = 1$:
$1 = -\frac{1^3}{2} + C(1)$
$C = 1 + \frac{1}{2} = \frac{3}{2}$

Thus, the function $f(x)$ is defined for all $x \in \mathbf{R}$ as:
$f(x) = -\frac{x^3}{2} + \frac{3}{2}x$

Now, we calculate the value of $f(-3)$:
$f(-3) = -\frac{(-3)^3}{2} + \frac{3}{2}(-3)$
$f(-3) = -\frac{-27}{2} - \frac{9}{2}$
$f(-3) = \frac{27 - 9}{2} = \frac{18}{2} = 9$