Find the point in the closed unit disc $D=\{ (x,y) | x^2+y^2\le 1 \}$ at which the function $f(x,y)=x+y$ attains its maximum .
1. **Identify the domain and function**: We are given the closed unit disc \( D = \{ (x,y) \mid x^2 + y^2 \le 1 \} \) and the function \( f(x,y) = x + y \). We need to find the point in \( D \) where \( f(x,y) \) attains its maximum.

2. **Analyze the boundary**: The boundary of the disc is the unit circle \( x^2 + y^2 = 1 \). The function \( f(x,y) = x + y \) represents a family of lines with slope \(-1\) (since \( y = -x + c \)).

3. **Find the points of tangency**: The maximum value of \( f(x,y) \) on the boundary will occur where the line \( x + y = c \) is tangent to the circle \( x^2 + y^2 = 1 \). At the point of tangency, the gradient of the circle (which is perpendicular to the radius) will be parallel to the gradient of the function \( f(x,y) \).

4. **Use geometry to find the points**: The line \( x + y = c \) is tangent to the circle when the distance from the origin to the line is 1 (the radius of the circle). The distance from the origin to the line \( x + y = c \) is given by:
   \[
   \frac{|c|}{\sqrt{1^2 + 1^2}} = \frac{|c|}{\sqrt{2}}
   \]
   Setting this equal to 1 (the radius of the circle), we get:
   \[
   \frac{|c|}{\sqrt{2}} = 1 \implies |c| = \sqrt{2}
   \]
   Therefore, \( c = \sqrt{2} \) or \( c = -\sqrt{2} \).

5. **Find the points of intersection**: The lines \( x + y = \sqrt{2} \) and \( x + y = -\sqrt{2} \) intersect the circle at:
   \[
   x + y = \sqrt{2} \quad \text{and} \quad x^2 + y^2 = 1
   \]
   Solving these simultaneously, we substitute \( y = \sqrt{2} - x \) into \( x^2 + y^2 = 1 \):
   \[
   x^2 + (\sqrt{2} - x)^2 = 1
   \]
   \[
   x^2 + 2 - 2\sqrt{2}x + x^2 = 1
   \]
   \[
   2x^2 - 2\sqrt{2}x + 2 = 1
   \]
   \[
   2x^2 - 2\sqrt{2}x + 1 = 0
   \]
   Solving this quadratic equation:
   \[
   x = \frac{2\sqrt{2} \pm \sqrt{(2\sqrt{2})^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2}
   \]
   \[
   x = \frac{2\sqrt{2} \pm \sqrt{8 - 8}}{4}
   \]
   \[
   x = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}
   \]
   Thus, \( x = \frac{1}{\sqrt{2}} \) and \( y = \frac{1}{\sqrt{2}} \).

6. **Verify the points**: Similarly, for \( x + y = -\sqrt{2} \), we get the points \( x = -\frac{1}{\sqrt{2}} \) and \( y = -\frac{1}{\sqrt{2}} \).

7. **Determine maximum and minimum**: Evaluating \( f(x,y) \) at these points:
   \[
   f\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}
   \]
   \[
   f\left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}
   \]
   Therefore, the maximum value of \( f(x,y) \) is \( \sqrt{2} \) at the point \( \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \).

The final answer is \( \boxed{ \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) } \).