(a) To find the vertical asymptote, set the denominator equal to zero: \(x - 3 = 0\), so \(x = 3\).
For the oblique asymptote, perform polynomial long division on \(\frac{x^2}{x-3}\):

\(x^2 \div (x-3) = x + 3 + \frac{9}{x-3}\).
As \(x \to \infty\), \(\frac{9}{x-3} \to 0\), so the oblique asymptote is \(y = x + 3\).

(b) Substitute \(y = \frac{x^2}{x-3}\) into \(yx - 3y = x^2\) to form a quadratic: \(x^2 - yx + 3y = 0\).
The discriminant \(b^2 - 4ac = y^2 - 4(3y) = y^2 - 12y\) must be negative for no real solutions.
\(y^2 - 12y < 0\) implies \(0 < y < 12\), showing no points exist for \(0 < y < 12\).

(c) Sketch the curve showing the vertical asymptote at \(x = 3\) and the oblique asymptote \(y = x + 3\). The curve approaches these asymptotes.

(d)(i) Sketch \(y = \left| \frac{x^2}{x-3} \right|\) and \(y = |x| - 3\).
Find intercepts: \(y = \left| \frac{x^2}{x-3} \right|\) intersects the y-axis at \((0,0)\).
\(y = |x| - 3\) intersects the y-axis at \((0,-3)\) and x-axis at \((3,0)\) and \((-3,0)\).

(d)(ii) From the sketch, \(\left| \frac{x^2}{x-3} \right| \leq |x| + c\) has no solution when \(c < -3\).