Find all real-valued functions $f$ on the reals such that $f(-x) = -f(x)$, $f(x+1) = f(x) + 1$ for all $x$, and $f\left(\dfrac{1}{x}\right) = \dfrac{f(x)}{x^2}$ for $x \not = 0$.
1. **Given Conditions:**
   - \( f(-x) = -f(x) \) for all \( x \in \mathbb{R} \)
   - \( f(x+1) = f(x) + 1 \) for all \( x \in \mathbb{R} \)
   - \( f\left(\frac{1}{x}\right) = \frac{f(x)}{x^2} \) for \( x \neq 0 \)

2. **Evaluate at \( x = 0 \):**
   - From \( f(-x) = -f(x) \), set \( x = 0 \):
     \[
     f(0) = -f(0) \implies 2f(0) = 0 \implies f(0) = 0
     \]

3. **Define a new function \( g(x) \):**
   - Let \( g(x) = f(x) - x \)
   - Then, we have:
     \[
     f(x) = g(x) + x
     \]

4. **Transform the given conditions using \( g(x) \):**
   - From \( f(-x) = -f(x) \):
     \[
     g(-x) + (-x) = -[g(x) + x] \implies g(-x) - x = -g(x) - x \implies g(-x) = -g(x)
     \]
   - From \( f(x+1) = f(x) + 1 \):
     \[
     g(x+1) + (x+1) = g(x) + x + 1 \implies g(x+1) + x + 1 = g(x) + x + 1 \implies g(x+1) = g(x)
     \]
   - From \( f\left(\frac{1}{x}\right) = \frac{f(x)}{x^2} \):
     \[
     g\left(\frac{1}{x}\right) + \frac{1}{x} = \frac{g(x) + x}{x^2} \implies g\left(\frac{1}{x}\right) + \frac{1}{x} = \frac{g(x)}{x^2} + \frac{1}{x}
     \]
     \[
     g\left(\frac{1}{x}\right) = \frac{g(x)}{x^2}
     \]

5. **Simplify using periodicity and odd function properties:**
   - From \( g(x+1) = g(x) \), \( g(x) \) is periodic with period 1.
   - From \( g(-x) = -g(x) \), \( g(x) \) is an odd function.

6. **Evaluate \( g \left( \frac{x}{x+1} \) and \( g \left( \frac{x+1}{x} \) using the properties:**
   - From \( g\left(\frac{1}{x+1}\right) = \frac{g(x+1)}{(x+1)^2} \):
     \[
     g\left(\frac{1}{x+1}\right) = \frac{g(x)}{(x+1)^2}
     \]
   - From \( g(1-x) = -g(x) \):
     \[
     g\left(\frac{x}{x+1}\right) = g\left(1 - \frac{1}{x+1}\right) = -g\left(\frac{1}{x+1}\right) = \frac{-g(x)}{(x+1)^2}
     \]
   - From \( g\left(\frac{x+1}{x}\right) = g\left(1 + \frac{1}{x}\right) = g\left(\frac{1}{x}\right) \):
     \[
     g\left(\frac{x+1}{x}\right) = \frac{g(x)}{x^2}
     \]

7. **Equate and solve for \( g(x) \):**
   - From \( g\left(\frac{x+1}{x}\right) = \frac{-g(x)}{x^2} \):
     \[
     \frac{-g(x)}{x^2} = \frac{g(x)}{x^2} \implies -g(x) = g(x) \implies 2g(x) = 0 \implies g(x) = 0
     \]

8. **Conclude the function \( f(x) \):**
   - Since \( g(x) = 0 \), we have:
     \[
     f(x) = g(x) + x = x
     \]

The final answer is \( \boxed{f(x) = x} \).