Find the absolute extrema of 
F(x) = 1/3 x² - 2x² + 3x -        [-2, 5]

Find dy/dx:  xy + x +y = x²y²

Find ds/dt: s³ + st +4t³ -8t = 0

Find the equation of the line tangent to the curve at the given point;
X^¾ + y^¾ = 2    (1, 1)

Evaluate dy/dt at the given point:
x²e^y -y²ln x = 1 dx/dt = 1, x = 1, y =3

Well, here is the solution for one of your problems doing implicit
differentiation. From here, you should be able to solve it. You also
might want to check out this website: http://www.calc101.com . It is a
GREAT website when it comes to help. It shows you how to differentiate
functions when you're having problems and especially when you get to
integrals (if you haven't already). I am a math student myself so of
course I might be incorrect in some of the steps that I am showing
you, but for the most part I believe it is correct. Also another note,
I am showing you all the steps involved with solving this. I am not
doing this to patronize you or to mock you, I am just doing this to
show you every little single step from looking at the problem to
solving it.

When you do implicit differentiation, you need to look at each
separate function. From there, you need to see what you are actually
differentiating as far as respect to what variable (for instance x or
y). So, if you notice the first part (x*y) you will have to treat
those as two separate functions. So of course, you need to use the
product rule. So to take the dy/dx of that, you do this:   f'(x)*y +
f'(y)*x  so then you would have to do something like this:  dy/dx(x) *
y + dy/dx(y) *x.  This of course leaves you with something like this: 
 y + dy/dx*x . This is because for the first part, if you take the
derivative of y with respect to x OF the function x, it is just 1, and
then you multiply it times y (because of the product rule)  that is
the first part, and then the 2nd part is, you take the derivative of y
with respect to x OF the function y =   1 * (dy/dx) becuase you're
taking the derivative of the y with respect to x AND then multiplying
it by x (once again because of the product rule). And the for the
remainder you just follow the same steps, so dy/dx(x) = 1    and then 
dy/dx(y) = 1 * (dy/dx) . And then of course you have the equals sign
and on the right part, you have another (guess again) product rule
because you have 2 functions being multiplied with each other. so then
it looks similar to the first part. so its   dy/dx(x^2) * y^2  + 
dy/dx(y^2) * x^2. so to evaluate that, you do this   dy/dx(x^2) =  2*x
(becuase you just take a regular derivative of that function) and then
you multiply it with y^2 (product rule again). Then, once again
following the product rule (learn these rules because they come back
to haunt you throughout your calculus studies!!) you do the other part
 so its   dy/dx(y^2)*x^2  so evaluating that, give you 
2*y(dy/dx)*x^2. When you do implicit differentiation, the term (dy/dx)
will be like another variable within your function. So from there, you
need to group all the (dy/dx) and then solve for that variable. so
then you get something like this:    
y+1-2*x*y^2=[2*y*x^2*(dy/dx)]-[(dy/dx)*x]-[(dy/dx)]. so your final
answer is    dy/dx = [(y+1-2*x*y^2)/(2x^2*y-x+1)] . Also just another
note, looking at the 5th problem. If you want to evaluate dy/dx at a
given point, all you do is solve for the dy/dx and then AT the end,
you just plug in what x is and what y is and then you get an answer
from there.

Hi, it's me again. Well I kinda didn't enter in the FINAL answer
properly. Here it is:  dy/dx = [(y+1-2*x*y^2)/(2x^2*y-x-1)]   I put a
+ sign instead of the - sign for the denominator. That should clear
things up.