Let \(f\) be a real-valued differentiable function on \(R\) (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 the value of \(f(-3)\) is equal to

The equation of the tangent at \((x, y)\) to the given curve \(y=f(x)\) is
\[
\begin{gathered}
Y-y=\frac{d y}{d x}(X-x) \\
Y \text {-intercept }=y-x \frac{d y}{d x}
\end{gathered}
\]
According to the question
\[
x^3=y-x \frac{d y}{d x}
\]

\[
\Rightarrow \quad \frac{d y}{d x}-\frac{y}{x}=-x^2
\]
which is linear in \(x\).
\[
\text { IF }=e^{\int \frac{-1}{x} d x}=\frac{1}{x}
\]
\(\therefore\) Required solution is
\[
\begin{aligned}
& y \cdot \frac{1}{x}=\int-x^2 \cdot \frac{1}{x} d x \\
& \Rightarrow \quad \frac{y}{x}=\frac{-x^2}{2}+c \\
& \Rightarrow \quad y=\frac{-x^3}{2}+c x \\
& \text { At } x=1, y=1 \text {, } \\
& 1=\frac{-1}{2}+c \\
& \Rightarrow \quad c=\frac{3}{2} \\
&
\end{aligned}
\]
Now, \(\begin{aligned} f(-3) & =\frac{27}{2}+\frac{3}{2}(-3) \\ & =\frac{27-9}{2}=9\end{aligned}\)