Suppose `A=(dy)/(dx)` of `x^(2)+y^(2)=4` at `(sqrt2,sqrt2), B=(dy)/(dx)` of `sin y+sinx=sinx.siny` at `(pi,pi)andC=(dy)/(dx)` of `2e^(xy)+e^(x).e^(y)-e^(x)-e^(y)=e^(xy+1)` at `(1,1),` then `(A-B-C)` has the value equal to.......

Answer

To solve the problem, we need to find the values of \( A \), \( B \), and \( C \) which are the derivatives \( \frac{dy}{dx} \) at specific points for the given equations. Finally, we will compute \( A - B - C \).

### Step 1: Calculate \( A \)
We start with the equation:
\[ x^2 + y^2 = 4 \]
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4) \]
This gives:
\[ 2x + 2y \frac{dy}{dx} = 0 \]
Now, we can solve for \( \frac{dy}{dx} \):
\[ 2y \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y} \]
Now we substitute \( x = \sqrt{2} \) and \( y = \sqrt{2} \):
\[ \frac{dy}{dx} = -\frac{\sqrt{2}}{\sqrt{2}} = -1 \]
Thus, \( A = -1 \).

### Step 2: Calculate \( B \)
Next, we consider the equation:
\[ \sin y + \sin x = \sin x \sin y \]
Differentiating both sides with respect to \( x \):
\[ \cos y \frac{dy}{dx} + \cos x = \sin y \cos x \frac{dy}{dx} + \sin x \cos y \frac{dy}{dx} \]
Rearranging gives:
\[ \cos y \frac{dy}{dx} - \sin y \cos x \frac{dy}{dx} = -\cos x \]
Factoring out \( \frac{dy}{dx} \):
\[ \left( \cos y - \sin y \cos x \right) \frac{dy}{dx} = -\cos x \]
Now substituting \( x = \pi \) and \( y = \pi \):
\[ \cos \pi = -1, \quad \sin \pi = 0 \]
Thus, we have:
\[ \left( -1 - 0 \right) \frac{dy}{dx} = 1 \implies -\frac{dy}{dx} = 1 \implies \frac{dy}{dx} = -1 \]
So, \( B = -1 \).

### Step 3: Calculate \( C \)
Now we consider the equation:
\[ 2e^{xy} + e^x e^y - e^x - e^y = e^{xy + 1} \]
Differentiating both sides with respect to \( x \):
\[ 2e^{xy}(y + x \frac{dy}{dx}) + e^x e^y \frac{dy}{dx} - e^x \frac{dy}{dx} - e^y \frac{dy}{dx} = e^{xy + 1}(y + x \frac{dy}{dx}) \]
Now substituting \( x = 1 \) and \( y = 1 \):
\[ 2e^{1}(1 + 1 \frac{dy}{dx}) + e^1 e^1 \frac{dy}{dx} - e^1 \frac{dy}{dx} - e^1 \frac{dy}{dx} = e^{1 + 1}(1 + 1 \frac{dy}{dx}) \]
This simplifies to:
\[ 2e(1 + \frac{dy}{dx}) + e^2 \frac{dy}{dx} - e \frac{dy}{dx} - e \frac{dy}{dx} = e^2(1 + \frac{dy}{dx}) \]
Combining terms:
\[ 2e + 2e \frac{dy}{dx} = e^2 + e^2 \frac{dy}{dx} \]
Rearranging gives:
\[ (2e - e^2) + (2e - e^2) \frac{dy}{dx} = 0 \]
Thus, we find:
\[ \frac{dy}{dx} = \frac{2e}{e^2 - 2e} = -1 \]
So, \( C = -1 \).

### Final Calculation
Now we compute \( A - B - C \):
\[ A - B - C = -1 - (-1) - (-1) = -1 + 1 + 1 = 1 \]

### Conclusion
The final value of \( A - B - C \) is:
\[ \boxed{1} \]