`f(x) = {{:((sqrt(1+kx)-sqrt(1-kx))/(x),if -1 le x lt 0),((2x+1)/(x-1),if 0 le x le 1):}` at `x = 0`.

Explanation

To determine the value of \( k \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit as \( x \) approaches 0 from the negative side equals the right-hand limit as \( x \) approaches 0 from the positive side, and both equal \( f(0) \).

### Step 1: Find \( f(0) \) from the right side
For \( 0 \leq x \leq 1 \), the function is defined as:
\[ f(x) = \frac{2x + 1}{x - 1} \]
Now, substituting \( x = 0 \):
\[ f(0) = \frac{2(0) + 1}{0 - 1} = \frac{1}{-1} = -1 \]

### Step 2: Find the left-hand limit as \( x \) approaches 0
For \( -1 \leq x < 0 \), the function is defined as:
\[ f(x) = \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x} \]
We need to find:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x} \]

### Step 3: Rationalize the numerator
To simplify the limit, we can multiply the numerator and the denominator by the conjugate of the numerator:
\[ \lim_{x \to 0^-} \frac{(\sqrt{1 + kx} - \sqrt{1 - kx})(\sqrt{1 + kx} + \sqrt{1 - kx})}{x(\sqrt{1 + kx} + \sqrt{1 - kx})} \]
This gives:
\[ = \lim_{x \to 0^-} \frac{(1 + kx) - (1 - kx)}{x(\sqrt{1 + kx} + \sqrt{1 - kx})} \]
\[ = \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1 + kx} + \sqrt{1 - kx})} \]
\[ = \lim_{x \to 0^-} \frac{2k}{\sqrt{1 + kx} + \sqrt{1 - kx}} \]

### Step 4: Evaluate the limit
As \( x \to 0 \), \( \sqrt{1 + kx} \to 1 \) and \( \sqrt{1 - kx} \to 1 \):
\[ \lim_{x \to 0^-} f(x) = \frac{2k}{1 + 1} = \frac{2k}{2} = k \]

### Step 5: Set the limits equal for continuity
For \( f(x) \) to be continuous at \( x = 0 \):
\[ f(0) = \lim_{x \to 0^-} f(x) \]
This means:
\[ -1 = k \]

### Conclusion
Thus, the value of \( k \) that makes the function continuous at \( x = 0 \) is:
\[ \boxed{-1} \]