Finding the slopes of the tangent I am having trouble with the following part of this question. I have found the derivative of the function $2y^2-xy-x^2=0$:
$$\dfrac{dy}{dx}=\dfrac{2x+y}{4y-x}$$
But now I have to find the slopes of the tangents at $x=1$.
How do I go about doing this part? 
 A: First, it is important two note that this is not a function but a curve. That is because a function by definition cannot have more than one y value for a given x.
Start by solving the equation $2y^2-xy-x^2=0%$ at $x=1$ to get the values of y. You will get two values for y because it is a second order polynomial equation. This means that there are two points of the curve at $x=1$.
Excuse any mistakes, because it's pretty late here and I'm tired but I got $y_1 = 1$ and $y_2 = -1/2$. Now, solve the equation of the derivative for each of the values of y. Each of the two solutions corresponds to the tangent of the value of y you used of course. I leave the calculations to you as they are pretty straightforward.
A: You've done the hard part. 
At $x=1,$ we have $\color{green}{2y^2-1\cdot y-1^2=0} \quad$ (plugging $x=0$ into the original equation).
Solve $\color{green}{\text{this}}$ quadratic for $y,$ then find $\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{x=1}$ by subbing in $x=1$ and $y=y_1, y_2$ into your equation for $\frac{\mathrm{d}y}{\mathrm{d}x},$ where $y_1$ and $y_2$ are solutions to the above quadratic equation.
