The equation of the tangent at (x, y) to the given curve y = f(x) is

$$Y - y = {{dy} \over {dx}}(X - x)$$

Y-intercept $$ = y - x{{dy} \over {dx}}$$

According to the question

$${x^3} = y - x{{dy} \over {dx}}$$

$$ \Rightarrow {{dy} \over {dx}} - {y \over x} = - {x^2}$$

which is linear in x.

$$IF = {e^{\int {{{ - 1} \over x}dx} }} = {1 \over x}$$

$$\therefore$$ Required solution is

$$y\,.\,{1 \over x} = \int { - {x^2}\,.\,{1 \over x}dx} $$

$$ \Rightarrow {y \over x} = {{ - {x^2}} \over 2} + c$$

$$ \Rightarrow y = {{ - {x^3}} \over 2} + cx$$

At x = 1, y = 1,

$$1 = {{ - 1} \over 2} + c$$

$$ \Rightarrow c = {3 \over 2}$$

Now, $$f( - 3) = {{27} \over 2} + {3 \over 2}( - 3)$$

$$ = {{27 - 9} \over 2} = 9$$