By differentiation with respect to \(\beta \), we have \[ f'(\beta ) = 1 + \frac {1}{\beta ^2} + \frac {2}{\beta ^3}. \]
If \(f'(t) = 0\), we must have \[ t^3 + t + 2 = 0. \]
Therefore, \[ (t + 1)(t^2 - t + 2) = 0, \] and hence the only real root to this is \(t = -1\), since \((-1)^2 - 2 \cdot 4 < 0\).
This means the only stationary point of \(y = f(\beta )\) is \((-1, f(-1) = -1)\).
For the limiting behaviour of the function, we first look at the case where \(\beta > 0\). As \(\beta \to \infty \), we have \(f(\beta ) \to \beta \) from below. As \(\beta \to 0^{+}\), we have \(f(\beta ) \to -\frac {1}{\beta } - \frac {1}{\beta ^2} \to -\infty \).
When \(\beta < 0\), we use the substitution \(t = -\frac {1}{\beta }\) to make the behaviours more convincing, and hence \[ f(\beta ) = \beta + t - t^2. \]
As \(\beta \to 0^{-}\), we have \(t \to \infty \), and \(f(\beta ) \to t - t^2 \to -\infty \). As \(\beta \to -\infty \), we have \(t \to 0^{+}\), and \(f(\beta ) \to \beta \) from above, since \(t - t^2 = t (1 - t) > 0\) when \(0 < t < 1\).
This means the curve \(y = f(\beta )\) is as below.
Similarly, by differentiation with respect to \(\beta \), we have \[ g'(\beta ) = 1 - \frac {3}{\beta ^2} + \frac {2}{\beta ^3}. \]
If \(g'(t) = 0\), we must have \[ t^3 - 3t + 2 = 0. \]
Therefore, \[ (t - 1)^2(t + 2) = 0, \] and hence the real roots to this is \(t = 1\) and \(t = -2\).
This means the stationary points of \(y = g(\beta )\) is \((1, g(1) = 3)\) and \((-2, g(-2) = -\frac {15}{4})\).
For the limiting behaviour of the function, we first look at the case where \(\beta > 0\). We consider the substitution \(t = -\frac {1}{\beta }\) to make the behaviours more convincing, and hence \[ g(\beta ) = \beta - 3t - t^2. \]
As \(\beta \to \infty \), \(t \to 0^{-}\), and hence \(f(\beta ) \to \beta \) from below, since \(-3t - t^2 = - t(t + 3) > 0\) for \(-3 < t < 0\). As \(\beta \to 0^{+}\), \(t \to -\infty \), and hence \(f(\beta ) \to -3t - t^2 \to -\infty \).
When \(\beta < 0\), we have as \(\beta \to 0^{-}\), \(f(\beta ) \to -\infty \). As \(\beta \to -\infty \), \(f(\beta ) \to \beta \) from below.
This means the curve \(y = g(\beta )\) is as below.
By the given condition, we have \[ -\alpha + \frac {1}{\beta } = -1 \iff \alpha = 1 + \frac {1}{\beta }. \]
Hence, \begin {align*} \frac {1}{u} + \frac {1}{v} + uv & = - \frac {\alpha }{\beta } + \beta \\ & = - \frac {1 + \frac {1}{\beta }}{\beta } + \beta \\ & = \frac {\beta ^2 - 1 - \frac {1}{\beta }}{\beta } \\ & = \beta - \frac {1}{\beta } - \frac {1}{\beta ^2} \\ & = f(\beta ). \end {align*}
Also, since \(u, v\) are both real, we have \begin {align*} \alpha ^2 - 4\beta & = \left (1 + \frac {1}{\beta }\right )^2 - 4\beta \\ & = 1 + \frac {2}{\beta } + \frac {1}{\beta ^2} - 4\beta \\ & = \frac {-4 \beta ^3 + \beta ^2 + 2\beta + 1}{\beta ^2} \\ & \geq 0. \end {align*}
Multiplying both sides by \(-\beta ^2\) (which flips the sign) gives \begin {align*} 4\beta ^3 - \beta ^2 - 2\beta - 1 & \leq 0 \\ (\beta - 1)(4\beta ^2 + 3\beta + 1) & \leq 0. \end {align*}
This cubic has exactly one real root \(\beta = 1\), so the solution to this inequality is \(\beta \leq 1\) and \(\beta \neq 0\).
Notice that \(f\) is increasing on \((0, 1] \subset (0, \infty )\). Therefore, for \(\beta > 0\), \[ f(\beta ) \leq f(1) = 1 - 1 - 1 = -1. \]
When \(\beta < 0\), we have \[ f(\beta ) \leq f(-1) = -1. \]
So for the range of \(\beta \) in this question, we always have \(f(\beta ) \leq -1\). But we also have \(\frac {1}{u} + \frac {1}{v} + uv \leq -1\) as shown before. These gives us exactly our desired statement.
By the given condition, we have \[ -\alpha + \frac {1}{\beta } = 3 \iff \alpha = -3 + \frac {1}{\beta }. \]
Hence, \begin {align*} \frac {1}{u} + \frac {1}{v} + uv & = - \frac {\alpha }{\beta } + \beta \\ & = - \frac {-3 + \frac {1}{\beta }}{\beta } + \beta \\ & = \beta + \frac {3}{\beta } - \frac {1}{\beta ^2} \\ & = g(\beta ). \end {align*}
Also, since \(u, v\) are both real, we have \(\beta \leq 1\) and \(\beta \neq 0\) as well.
\(g\) must be increasing on \((0, 1]\). Hence, for \(\beta > 0\), we have \[ g(\beta ) \leq g(1) = 3. \]
When \(\beta < 0\), we have \[ g(\beta ) \leq g(-2) = -\frac {15}{4}. \]
Since \(3 > -\frac {15}{4}\), we can conclude that the maximum value of \(\frac {1}{u} + \frac {1}{v} + uv\) is \(3\), and it is taken when \(\beta = 1\), which corresponds to \(\alpha = -2\).