A curve has equation $y=f(x)$, where $f(x)=x^{3}+ax^{2}+bx+c$. The curve $y=f(x)$ has a maximum point at $(1,-32)$, while the curve $y=\dfrac{1}{f(x)}$ has a vertical asymptote with equation $x=5$. Find the equation of the curve $y=f(x)$.

1. Express the condition that the curve has a stationary point at $x=1$ as an equation in terms of $a$ and $b$.

$f'(x) = 3x^2 + 2ax + b$
At the stationary point $(1, -32)$, $f'(1) = 0$:
$3(1)^2 + 2a(1) + b = 0$
$2a + b = -3 \quad \text{--- (1)}$

2. Express the condition that the curve passes through the maximum point $(1, -32)$ as an equation in terms of $a, b$, and $c$.

Since the curve passes through $(1, -32)$, we have $f(1) = -32$:
$(1)^3 + a(1)^2 + b(1) + c = -32$
$1 + a + b + c = -32$
$a + b + c = -33 \quad \text{--- (2)}$

3. Relate the vertical asymptote of $y = \frac{1}{f(x)}$ to the roots of $f(x)$ to form a third equation.

For the curve $y = \frac{1}{f(x)}$ to have a vertical asymptote at $x = 5$, the denominator must be zero at this point. Hence, $f(5) = 0$.
$(5)^3 + a(5)^2 + b(5) + c = 0$
$125 + 25a + 5b + c = 0$
$25a + 5b + c = -125 \quad \text{--- (3)}$

4. Solve the system of linear equations simultaneously to find $a, b$, and $c$, and state the equation of the curve.

From (1), we can express $b$ in terms of $a$:
$b = -3 - 2a$
Substitute $b$ into (2):
$a + (-3 - 2a) + c = -33$
$-a - 3 + c = -33 \implies c = a - 30$
Substitute both $b$ and $c$ into (3):
$25a + 5(-3 - 2a) + (a - 30) = -125$
$25a - 15 - 10a + a - 30 = -125$
$16a - 45 = -125$
$16a = -80 \implies a = -5$
Substitute $a = -5$ back into the expressions for $b$ and $c$:
$b = -3 - 2(-5) = -3 + 10 = 7$
$c = -5 - 30 = -35$
Thus, the equation of the curve is $y = x^3 - 5x^2 + 7x - 35$.