Suppose a continuous function $f:[0,1]\to\mathbb{R}$ is differentiable on $(0,1)$ and $f(0)=1$, $f(1)=0$. Then, there exists $0<x_0<1$ such that
$$|f'(x_0)| \geq 2018 f(x_0)^{2018}$$
1. **Initial Assumptions and Definitions:**
   - Given a continuous function \( f: [0,1] \to \mathbb{R} \) that is differentiable on \( (0,1) \).
   - Boundary conditions: \( f(0) = 1 \) and \( f(1) = 0 \).
   - We need to show that there exists \( 0 < x_0 < 1 \) such that \( |f'(x_0)| \geq 2018 f(x_0)^{2018} \).

2. **Case Analysis:**
   - If there exists \( x_0 \in (0,1) \) such that \( f(x_0) = 0 \), then \( |f'(x_0)| \geq 2018 f(x_0)^{2018} \) holds trivially because \( 0 \geq 0 \).

3. **Assume \( f(x) \neq 0 \) for \( x \in (0,1) \):**
   - By the Intermediate Value Theorem, since \( f \) is continuous and \( f(0) = 1 \) and \( f(1) = 0 \), \( f(x) > 0 \) for all \( x \in [0,1) \).

4. **Define a New Function \( g \):**
   - Define \( g: [0,1) \to \mathbb{R} \) by \( g(x) = \frac{1}{f(x)^{2017}} \).
   - Since \( f \) is continuous at \( 1 \) and \( f(1) = 0 \), \( \lim_{x \to 1} g(x) = \infty \).

5. **Existence of \( x_1 \):**
   - There exists \( x_1 \in (0,1) \) such that \( g(x_1) = 2017 \cdot 2018 + 1 \).

6. **Application of the Mean Value Theorem:**
   - By the Mean Value Theorem, there exists \( x_0 \in (0, x_1) \) such that
     \[
     g'(x_0) = \frac{g(x_1) - g(0)}{x_1 - 0}.
     \]

7. **Compute \( g'(x) \):**
   - Since \( g(x) = \frac{1}{f(x)^{2017}} \), we have
     \[
     g'(x) = -2017 \cdot \frac{f'(x)}{f(x)^{2018}}.
     \]
   - Therefore,
     \[
     \frac{-2017 \cdot f'(x_0)}{f(x_0)^{2018}} = \frac{g(x_1) - g(0)}{x_1 - 0}.
     \]

8. **Estimate \( g(x_1) - g(0) \):**
   - Since \( g(x_1) = 2017 \cdot 2018 + 1 \) and \( g(0) = 1 \),
     \[
     g(x_1) - g(0) = 2017 \cdot 2018.
     \]
   - Thus,
     \[
     \frac{-2017 \cdot f'(x_0)}{f(x_0)^{2018}} = \frac{2017 \cdot 2018}{x_1}.
     \]

9. **Conclusion:**
   - Since \( x_1 \in (0,1) \), we have
     \[
     \frac{2017 \cdot 2018}{x_1} > 2017 \cdot 2018.
     \]
   - Therefore,
     \[
     -2017 \cdot \frac{f'(x_0)}{f(x_0)^{2018}} > 2017 \cdot 2018.
     \]
   - This implies
     \[
     -f'(x_0) > 2018 \cdot f(x_0)^{2018}.
     \]
   - Hence,
     \[
     |f'(x_0)| \geq 2018 \cdot f(x_0)^{2018}.
     \]

The final answer is \( \boxed{ |f'(x_0)| \geq 2018 f(x_0)^{2018} } \).