For your first question, consider it written as:

y + (x^2)(y^3) = 10.

For the derivative, I got y' = -2x(y^3)/((3(x^2)(y^2) + 1))

You want to find f'(1), so you'll need to substitute x=1 into your derivative. Since you did implicit differentiation, though, you also need to have a value to substitute in for y. For that, use the fact that your original function passes through (1,2) (since f(1) = 2). You're finding the derivative of your function at that point, so substitute x = 1 and y = 2 into your derivative. That's it.

For your second question, the one you labeled "1.", find the derivative just like you did in the first one, implicitly.

I get:

2x + 2yy' = 2(2x^2 + 2y^2 - x)(4x + 4yy' -1) [using the chain rule on the right]

Now, plug in your x and y values (0 and 1/2, resp.) and then solve for y'. Remember that your derivative, y', gives you the slope of the tangent line to your function at a particular point, in this case (0, 1/2). So now you know the slope of the tangent line (f'(0), the y' from above) and you also know a point on the tangent line (0, 1/2). Since you know a point and the slope, use point-slope form to write the equation of the line.

Point-slope form:

(y - y_1) = m (x - x_1).

Use the coordinates of your point for (x_1, y_1) and the slope for m.

Hope this helps - I'll leave the last few for you. Once you find the derivatives, the idea behind finding the equation of the tangent line is the same - use point-slope form.