Differential equations is fourth semester calculus, so don't worry about it. The language makes it seem a lot worse than it really is.
That's so weird that your class did that bad. Implicit differentiation isn't that hard, and I always know if a test didn't go well.
Though, once I took multivariable calculus (that is, third semester), I learned this really neat thing called the Implicit Function Theorem. If all you want to do is find dy/dx, it makes things a lot simpler if you are lazy like me. Basically, if we make the function of the form F(x,y) = 0 (so we'll just put all the terms on one side), then
dy/dx = - (part(F)/part(x)) / (part(F)/part(y))
Those are partial derivatives up there, which are really easy to compute. If I wanted to take the partial derivative with respect to x of (y*x^3 + y^2*e^y), that's kind of a nasty thing to differentiate, and if you're doing implicit differentiation you might not even know how to take the derivative of transcendentals like e^y. Luckily, it doesn't matter at all. The "partial" derivative is when we pretend all other variables (except the one we are differentiating--in this case, x) are constants. Now the other variables are cake, because everyone can differentiate a constant. In that example I popped up there, that derivative is 3yx^2, because that whole other term has all y's in it, so it's effectively a giant constant, making it zero. The y will act like a constant in the other one, too. All differentiation things still apply.
I guess now that I think about it you could get some really ugly implicit differentiation on tests, especially if you have to apply the quotient rule with the product rule. It gets annoying quickly. So let's just skip that crap and hope our partial derivatives simplify things!
So take a partial derivative with respect to X of that function, then y. Your answer will be -dF/dx / dF/dy. (In reality, partial derivatives are denoted with a script d, rather than a d like I'm using, but heck if I can find script d for a forum... Let's pretend my notation is right.
When you think about it, it makes sense. When you are implicitly differentiating you'll get a whole bunch of junk like this:
dy / dx (junk of a function) = (different junk)
And to solve it, we divide the junks so we get dy/dx = junk/different junk. So while I'm not going to prove the Implicit Function Theorem (because that would be unpleasant and involve the Jacobian, which I don't feel like dealing with) hopefully you can sort of see from inspection that whenever you get an answer for your implicit differentiation, it will be the quotient of two functions. In effect, what you do with implicit differentiation is exactly the same as computing partials.
So that probably flew over your head. It might impress your teacher when you start whipping out the partials though.
(Indeed, any solution you get implicitly, you could check with this method. Perhaps it would be a good thing to remember for the final, to check your work. Hopefully you have time to do so, and if not, that's cruel.)
You might not believe me, but multivariable calculus was the easiest of all four semesters of calc, also the most fun. Differential equations isn't hard, either, as it basically boils down to "I know the type of solution it looks like, so here's the answer". There's plenty of ugly algebra though. Multivariable sounds ominous, but only once did I have truly atrocious algebra.
...I'm a little bored, as you can tell.
