Find all continuously differentiable functions $f:[0,1]\to(0,\infty)$ such that $\frac{f(1)}{f(0)}=e$ and
$$\int^1_0\frac{\text dx}{f(x)^2}+\int^1_0f'(x)^2\text dx\le2.$$
1. We start with the given conditions:
   \[
   \frac{f(1)}{f(0)} = e
   \]
   and
   \[
   \int_0^1 \frac{dx}{f(x)^2} + \int_0^1 f'(x)^2 \, dx \leq 2.
   \]

2. Consider the expression:
   \[
   0 \leq \int_0^1 \left( f'(x) - \frac{1}{f(x)} \right)^2 dx.
   \]
   Expanding the square, we get:
   \[
   \int_0^1 \left( f'(x) - \frac{1}{f(x)} \right)^2 dx = \int_0^1 f'(x)^2 \, dx - 2 \int_0^1 \frac{f'(x)}{f(x)} \, dx + \int_0^1 \frac{dx}{f(x)^2}.
   \]

3. Using the given inequality:
   \[
   \int_0^1 \frac{dx}{f(x)^2} + \int_0^1 f'(x)^2 \, dx \leq 2,
   \]
   we substitute into the expanded square:
   \[
   \int_0^1 f'(x)^2 \, dx - 2 \int_0^1 \frac{f'(x)}{f(x)} \, dx + \int_0^1 \frac{dx}{f(x)^2} \leq 2 - 2 \int_0^1 \frac{f'(x)}{f(x)} \, dx.
   \]

4. Notice that:
   \[
   \int_0^1 \frac{f'(x)}{f(x)} \, dx = \ln \left| f(x) \right| \bigg|_0^1 = \ln \frac{f(1)}{f(0)} = \ln e = 1.
   \]
   Therefore:
   \[
   \int_0^1 f'(x)^2 \, dx - 2 \cdot 1 + \int_0^1 \frac{dx}{f(x)^2} \leq 2 - 2,
   \]
   which simplifies to:
   \[
   \int_0^1 f'(x)^2 \, dx + \int_0^1 \frac{dx}{f(x)^2} - 2 \leq 0.
   \]

5. Since both integrals are non-negative, the only way the inequality can hold is if:
   \[
   \int_0^1 \left( f'(x) - \frac{1}{f(x)} \right)^2 dx = 0.
   \]
   This implies:
   \[
   f'(x) = \frac{1}{f(x)} \quad \forall x \in [0,1].
   \]

6. Solving the differential equation \( f'(x) = \frac{1}{f(x)} \):
   \[
   f(x) f'(x) = 1.
   \]
   Integrating both sides with respect to \( x \):
   \[
   \int f(x) f'(x) \, dx = \int 1 \, dx,
   \]
   which gives:
   \[
   \frac{1}{2} f(x)^2 = x + C,
   \]
   or:
   \[
   f(x)^2 = 2x + 2C.
   \]
   Therefore:
   \[
   f(x) = \sqrt{2x + 2C}.
   \]

7. Using the initial condition \( \frac{f(1)}{f(0)} = e \):
   \[
   \frac{\sqrt{2 \cdot 1 + 2C}}{\sqrt{2 \cdot 0 + 2C}} = e,
   \]
   which simplifies to:
   \[
   \frac{\sqrt{2 + 2C}}{\sqrt{2C}} = e.
   \]
   Squaring both sides:
   \[
   \frac{2 + 2C}{2C} = e^2,
   \]
   which simplifies to:
   \[
   1 + \frac{1}{C} = e^2.
   \]
   Solving for \( C \):
   \[
   \frac{1}{C} = e^2 - 1,
   \]
   so:
   \[
   C = \frac{1}{e^2 - 1}.
   \]

8. Therefore, the function \( f(x) \) is:
   \[
   f(x) = \sqrt{2x + \frac{2}{e^2 - 1}}.
   \]

The final answer is \(\boxed{f(x) = \sqrt{2x + \frac{2}{e^2 - 1}}}\).