Step 1: Formulate the differential equation from the given information.

Let the curve be $y = f(x)$. The tangent is drawn at an arbitrary point $P(x, y)$ on this curve.

The slope of the tangent at point $P(x,y)$ is given by the derivative of the function, $m = f'(x) = \frac{dy}{dx}$.

The equation of the tangent line at the point $P(x, y)$ can be written using the point-slope form: $Y - y = m(X - x)$ $Y - y = f'(x)(X - x)$ where $(X, Y)$ are the coordinates of any point on the tangent line.

To find the y-intercept of this tangent line, we set $X = 0$ in the equation: $Y_{intercept} - y = f'(x)(0 - x)$ $Y_{intercept} = y - x f'(x)$

The problem states that the y-intercept of the tangent is equal to the cube of the abscissa of point $P$. The abscissa of $P(x, y)$ is $x$. So, we have the condition: $Y_{intercept} = x^3$

Equating the two expressions for the y-intercept, we get the differential equation: $y - x f'(x) = x^3$ $y - x \frac{dy}{dx} = x^3$

Step 2: Solve the differential equation.

Rearrange the differential equation into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$. $-x \frac{dy}{dx} + y = x^3$ $x \frac{dy}{dx} - y = -x^3$ Assuming $x \neq 0$, we divide by $x$: $\frac{dy}{dx} - \frac{1}{x} y = -x^2$ This is a linear first-order differential equation with $P(x) = -\frac{1}{x}$ and $Q(x) = -x^2$.

We find the integrating factor (I.F.): $I.F. = e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = e^{\ln(|x|^{-1})} = \frac{1}{|x|}$ For simplicity, let's work with $x > 0$, so I.F. = $1/x$. The resulting function will be checked for validity for all real numbers.

Multiply the differential equation by the integrating factor: $\frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = -x$ The left side is the derivative of $(y \cdot I.F.)$: $\frac{d}{dx} \left( y \cdot \frac{1}{x} \right) = -x$

Now, integrate both sides with respect to $x$: $\int \frac{d}{dx} \left( \frac{y}{x} \right) dx = \int -x dx$ $\frac{y}{x} = -\frac{x^2}{2} + C$ where $C$ is the constant of integration.

Solving for $y$, we get the general solution for the function $f(x)$: $y = f(x) = -\frac{x^3}{2} + Cx$

Step 3: Use the initial condition to find the constant C.

We are given that $f(1) = 1$. Substitute $x=1$ and $y=1$ into the general solution: $1 = -\frac{(1)^3}{2} + C(1)$ $1 = -\frac{1}{2} + C$ $C = 1 + \frac{1}{2} = \frac{3}{2}$

Step 4: Determine the specific function f(x).

Substitute the value of $C$ back into the general solution: $f(x) = -\frac{x^3}{2} + \frac{3}{2}x$ This function is a polynomial, so it is differentiable on all of $R$, satisfying the problem's condition.

Step 5: Calculate the value of f(-3).

Finally, we need to find the value of $f(-3)$. Substitute $x = -3$ into the function: $f(-3) = -\frac{(-3)^3}{2} + \frac{3}{2}(-3)$ $f(-3) = -\frac{-27}{2} - \frac{9}{2}$ $f(-3) = \frac{27}{2} - \frac{9}{2}$ $f(-3) = \frac{27 - 9}{2}$ $f(-3) = \frac{18}{2}$ $f(-3) = 9$

Thus, the value of $f(-3)$ is 9.