Given the equation:

x^3 + y^3 = 14xy

We need to find the value of the first derivative at the point (7,7).

First we will use implicit differentiation.

==> (x^3)' + (y^3)' = 14(xy)'

==> 3x^2 + 3y^2 *y' = 14[ (x'y+xy')]

==> 3x^2 + 3y^2 y' = 14(y + xy')

==> 3x^2 + 3y^2 y' = 14y + 14xy'

Now we will combine terms with y'.

==> 3y^2 y' - 14xy' = 14y - 3x^2

==> y' ( 3y^2-14x) = 14y - 3x^2

==> y' = (14y-3x^2)/ (3y^2-14x)

Now we will substitute with (7,7)

==> y'(7,7) = (14*7 - 3*7^2) / (3*7^2 - 14*7)

= 98-147 / 147-98

= -49/49 = -1

**Then the value of the derivative is (-1).**

The expression to be differentiated is x^3 + y^3 = 14xy. This would require implicit differentiation.

x^3 + y^3 = 14xy

differentiating

3x^2 + 3*y^2(dy/dx) = 14x(dy/dx) + 14y

move all terms with dy/dx to one side

=> 3x^2 - 14y = (dy/dx)(14x - 3y^2)

=> dy/dx = (3x^2 - 14y)/(14x - 3y^2)

We require the value of the derivative at the point (7,7)

Substitute the values in the expression for the derivative dy/dx which we have arrived at.

dy/dx = (3*7^2 - 14*7)/(14*7 - 3*7^2)

=> (3*49 - 98)/(98 - 147)

=> -1

**The value of the derivative of x^3 + y^3 = 14xy at (7,7) is -1.**