If $x^2-xy+2y^2=\frac{a^2}{7}$, find $y'''$. 
If $x^2-xy+2y^2=\frac{a^2}{7}$, find $y'''$.

For our 1st derivative we got $$y'=\frac{2x-y}{x-4y}.$$
For the second derivative we got $$y''=\frac{14x^2-14xy+28y^2}{(x-4y)^3}.$$
And for the final answer we got $$y'''=\frac{4(-84x^3+119x^2y-154xy^2-84y^3)}{(x-4y)^5}.$$
Took me 2 hours and 3 white boards to get to that answer, and the teacher said another answer was right, also that there was an $a$ in the final answer, even though the derivative of a constant is zero. any insight or help is greatly appreciated, and will try to post all my work as an answer.

As Will Jagy noted, $$y''=\frac{2a^2}{(x-4y)^3}.$$
Then $$y'''=2a^2(-3)(x-4y)^{-4}(1-4y')$$ Now $$1-4y'=\frac{4y-8x-4y+x}{x-4y}=\frac{7x}{x-4y},$$ so your $y'''$ is $$y'''=\frac{42a^2x}{(x-4y)^5}.$$
I used your answer for second derivative to find the third derivative and I got $$y'''=\frac{(x-4y)(28x-14y-14xy'+56yy')-(14x^2-14xy+28y^2)\times 3(1-4y')}{(x-4y)^4}$$ after cancelling $(x-4y)^2$ out from numerator and denominator. Then I substitute $y'=\frac{2x-y}{x-4y}$ and simplify, and I got
$$\frac{294(x^2-xy+2y^2)x}{(x-4y)^5}$$ which agrees with my answer upon substituting $x^2-xy+2y^2=a^2/7$ . This means that if you had calculated correctly, your answer should have been $$y'''=\frac{294(x^2-xy+2y^2)x}{(x-4y)^5}.$$
