Implicit Differentation Did I tackle this implicit differentiation correctly?
$$5x^2+3xy-y^2=5$$
$$10x+3x\dfrac{dy}{dx}+3y-2y\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}(y-2y)=-10x-3z-3y$$
$$\dfrac{dy}{dx}=\dfrac{-10x-3x-3y}{y-2y}$$
 A: Beside to @Andre's answer, if you are familiar to partial differentiation then you can use the following rule: $$y'=-\frac{F_x(x,y)}{F_y(x,y)}$$ where in we assume our relation can be written as $F(x,y)=0$. Note that $y$ should be a function of $x$ at this rule.
A: Unfortunately not. There seems to be a mistake in your second line. Let me take you through the steps:
$$5x^2+3xy-y^2 = 5 \, ,$$
$$10x + 3y + 3x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0 \, ,$$
$$(3x-2y)\frac{dy}{dx} = -10x - 3y \, , $$
$$(2y-3x)\frac{dy}{dx} = 10x + 3y \, , $$
$$\frac{dy}{dx} = \frac{10x+3y}{2y-3x} \, . $$
Of course, this is not satisfactory. There should not be any $y$s on the right-hand-side. From the definition, we can obtain an expression for $y$:
$$y = \frac{3}{2}x \pm \frac{1}{2}\sqrt{29x^2-20} \, . $$
Putting this into the formula for $dy/dx$, we get:
$$\frac{dy}{dx} = \frac{\pm29x+3\sqrt{29x^2-20}}{\sqrt{29x^2-20}}$$
This is only well defined for $29x^2-20 > 0.$ When $29x^2-20=0$ the hyperbola will have vertical tangents, and so $dy/dx$ is not well-defined. When $29x^2-20 < 0$, there will be no corresponding $y$, i.e. you will be "inside" the hyperbola.
A: I am not sure where your $z$ is coming from. 
Steps 1 and 2 are correct,
but step 3 onwards, you group the terms incorrectly.
It should be $\frac {dy}{dx}(3x-2y)$
Therefore, it should be $\frac{dy}{dx}=\frac{-10x-3y}{3x-2y}$.
