The function f is defined by $f:x\mapsto\dfrac{1}{2-x^2},\quad x\in\mathbb{R},\;x\le 0,\;x\ne -\sqrt{2}\,.$

1. Set $y = f(x)$ and make $x$ the subject to find the rule for $f^{-1}$.

Let $y = \dfrac{1}{2-x^2}$.
$2 - x^2 = \dfrac{1}{y}$
$x^2 = 2 - \dfrac{1}{y}$
Since the domain of $f$ is $x \le 0$, we must take the negative square root:
$x = -\sqrt{2 - \dfrac{1}{y}}$

2. Determine the domain of $f^{-1}$ by finding the range of $f$.

The domain of $f^{-1}$ is the range of $f$.
Consider the two intervals in the domain of $f$ separated by the vertical asymptote $x = -\sqrt{2}$:
For $-\sqrt{2} < x \le 0$, $0 \le x^2 < 2 \implies 0 < 2-x^2 \le 2$. Thus, $f(x) \ge \dfrac{1}{2}$.
For $x < -\sqrt{2}$, $x^2 > 2 \implies 2-x^2 < 0$. Thus, $f(x) < 0$.
Hence, the range of $f$ is $y < 0 \text{ or } y \ge \dfrac{1}{2}$.

3. Express $f^{-1}$ in a form similar to the definition of $f$.

Combining the rule and the domain in the required notation:
$f^{-1}: x \mapsto -\sqrt{2 - \dfrac{1}{x}}, \quad x \in \mathbb{R}, \; x < 0 \text{ or } x \ge \dfrac{1}{2}$

1. State the asymptotes and sketch the graphs of $f$ and $f^{-1}$.

Exact equations of asymptotes:
For $f$: $x = -\sqrt{2}$ and $y = 0$
For $f^{-1}$: $x = 0$ and $y = -\sqrt{2}$

Sketch properties:

• The graphs of $f$ and $f^{-1}$ are reflections of each other about the line $y=x$ (drawn as a dashed line).
• $f$ has two branches: one passing through $(0, \frac{1}{2})$ with vertical asymptote $x=-\sqrt{2}$, and another in the third quadrant approaching $y=0$ as $x \to -\infty$.
• $f^{-1}$ has two branches: one passing through $(\frac{1}{2}, 0)$ with horizontal asymptote $y=-\sqrt{2}$, and another in the third quadrant approaching $x=0$ as $y \to -\infty$.

2. Set up the equation to find the intersection of $f$ and $f^{-1}$.

The graphs of a function and its inverse intersect on the line $y=x$ (for this relevant branch).
Setting $f(x) = x$:
$\frac{1}{2-x^2} = x$
$1 = 2x - x^3$
$x^3 - 2x + 1 = 0$

3. Solve the cubic equation for the exact $x$-coordinate of the intersection.

By the Factor Theorem or trial and error, $x=1$ is a root ($1^3 - 2(1) + 1 = 0$).
Factorising the cubic:
$(x-1)(x^2 + x - 1) = 0$
Using the quadratic formula for $x^2 + x - 1 = 0$:
$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$
Since the intersection must lie in the domain bounded by the asymptote ($x < -\sqrt{2}$), we reject $x=1$ and $x=\frac{-1+\sqrt{5}}{2}$.
Thus, the valid intersection point is at $x = \frac{-1-\sqrt{5}}{2}$.

4. Use the graphical sketch to find the values of x satisfying $f(x) \le f^{-1}(x)$.

From the sketch, we need the interval where the graph of $f$ lies on or below the graph of $f^{-1}$. Because $f$ and $f^{-1}$ are reflections about $y=x$ and are strictly decreasing, this inequality is satisfied when $f(x) \le x$.

From the graph, for the branch where $x < -\sqrt{2}$, $f(x)$ is bounded below $y=x$ starting from the intersection point and approaching the vertical asymptote.
For $x \in (-\sqrt{2}, 0)$, the graph of $f$ is positive while $f^{-1}$ is bounded below $y=-\sqrt{2}$, meaning $f(x) > f^{-1}(x)$ there.

Therefore, the required set of values is:
$\frac{-1-\sqrt{5}}{2} \le x < -\sqrt{2}$