= Solution
The <quotient rule> simplifies the <derivative> to
$$
y'=\frac{7(x+1)(x+5)}{(x-1)^2(x+2)^2}.
$$
Its zeros are $-5$ and $-1$, both in the domain. Substitution gives the two <stationary points>
$$
\boxed{(-5,25/9),\qquad(-1,9)}.
$$
The first is a <local maximum> and the second a <local minimum>, since the numerator of $y'$ changes sign from positive to negative at $-5$ and from negative to positive at $-1$. The denominator of the original <rational function> vanishes at
$$
\boxed{x=-2,\qquad x=1}.
$$
The numerator is nonzero at both values, so these discontinuities are poles, not removable holes.
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