We have two responses for you
Hi Sunny,
Let me try a similar expression,
(x^{2} + y^{2})^{2} = (x  y)^{2}
I want to find the slope of the tangent to this curve at (1, 1).
First I should check that the point is on the curve. At (1, 1) the left side is (1^{2} + (1)^{2})^{2} = 2^{2} = 4 and the right side is (1  (1))^{2} = 2^{2} = 4. Thus (1, 1) is on the curve.
The slope of the tangent to the curve at (1, 1) is the derivative of y with respect to x, evaluated at (1, 1). When I differentiate I need to remember that y is a function of x.
Differentiate the left side with respect to x.
Using the chain rule d/dx[(x^{2} + y^{2})^{2}] = 2(x^{2} + y^{2}) d/dx(x^{2} + y^{2})
I now need to differentiate (x^{2} + y^{2}) and I need to remember to use the chain rule again when I get to d/x(y^{2}). Thus d/x(y^{2}) = 2 y dy/dx. Hence putting it all together I get
d/dx[(x^{2} + y^{2})^{2}] = 2(x^{2} + y^{2}) d/dx(x^{2} + y^{2}) = 2(x^{2} + y^{2}) (2x + 2 y dy/dx)
Differentiate the right side with respect to x.
d/dx[(x  y)^{2}] = 2(x  y) d/dx(x  y) = 2(x  y) (1  dy/dx)
Since the left side is equal to the right side
2(x^{2} + y^{2}) (2x + 2 y dy/dx) = 2(x  y) (1  dy/dx)
To complete the problem I need to find dy/dx at (1, 1) so I substitute x = 1, y = 1 into the equation above to get
2(1 + 1) (2  2 dy/dx) = 2(1 + 1) (1  dy/dx)
2(2)(2  2 dy/dx) = 4(1  dy/dx)
divide both sides by 4 to get
2  2 dy/dx = 1  dy/dx
dy/dx = 1.
Now try your expression and write back if you still have difficulties.
Harley
 I'm assuming the "2" just before the equals sign is
an exponent, and have made the correction above.

To solve a problems like this, use implicit differentiation.
Let me illustrate with a simple example:
Find the slope of x^{4} + y^{6} = 1 + y^{2} at (1,1)
Take the derivative of each side with respect to x. When you hit a y, use the chain rule and leave the y' in place:
4x^{3} + 6 y^{5} y' = 2 y y'
Now isolate all the terms with y' on one side and factor out the y':
4 x^{3} = (2y  6y^{5}) y'
divide to isolate y' and plug in the coordinates to evaluate:
y' = 4 x^{3} /(2y  6y^{5}) = 1
 Implicit differentiation is also a very powerful technique in solving related rates problems and constrained optimization problems (It's not used as widely as it should be; too many students are nervous of it and try to eliminate a variable. Make it your friend  you'll be glad you did.) In applications, it's very helpful in thermodynamics, theoretical mechanics, and economics.
Good Hunting!
RD
