a) To find dy/dx, we can use implicit differentiation. Let's differentiate both sides of the equation with respect to x:
d/dx(xy + x^2y^2) = d/dx(x^3y^3)
Using the product rule on the first term, we get:
y + x(dy/dx) + 2xy^2 + 2x^2y(dy/dx) = 3x^2y^3 + 3x^3y^2(dy/dx)
Rearranging the terms, we have:
y + 2xy^2 = 3x^2y^3 - x(dy/dx) + 3x^3y^2(dy/dx) - 2x^2y(dy/dx)
Now, let's isolate dy/dx:
dy/dx(-x - 2x^2y + 3x^3y^2) = y + 2xy^2 - 3x^2y^3
dy/dx = (y + 2xy^2 - 3x^2y^3) / (-x - 2x^2y + 3x^3y^2)
b) We can apply implicit differentiation to find dy/dx for this equation as well. Differentiating both sides with respect to x:
d/dx(xsiny + ycosx) = d/dx(xy)
Using the product rule on the first term, we get:
siny + x(dy/dx)cosy - sinx + y(dy/dx)cosx = y + x(dy/dx)
Rearranging the terms, we have:
siny - sinx = x(dy/dx) - y(dy/dx)cosx + x(dy/dx)cosy - y
Now, let's isolate dy/dx:
dy/dx(x - ycosx + xcosy) = siny - sinx - y
dy/dx = (siny - sinx - y) / (x - ycosx + xcosy)