Implicit differentiation for $x^6-y^6=14$ I am trying to find the $\dfrac{\mathrm d^2 y}{\mathrm d x^2}$ for $x^6-y^6=14$.
I got $$\frac{5x^4\left(y^6-x^6\right)}{y^{11}}$$ but I'm not sure if it's right or not.
I am completely stuck on getting $\dfrac{\mathrm d^2 y}{\mathrm dx^2}$ for $y-\cos y =2x$.

It would be really helpful if you would also outline your computation. How can we see what your problem is when you do not tell about it?
In the first equation, differentiation of $x^6-y(x)^6=14$ gives
\begin{align}
6x^5-6y(x)^5y'(x)&=0\\
\implies y'(x)&=\frac{x^5}{y(x)^5}=\frac{x^5y(x)}{x^6-14}\\
5x^4-5y(x)^4y'(x)^2-y(x)^5y''(x)&=0\\
\implies
y''(x)&=\frac{5x^4}{y(x)^5}-\frac{5y'(x)^2}{y(x)}\\
&=\frac{5x^4}{y(x)^5}-\frac{5x^{10}}{y(x)^{11}}&
&=\frac{5x^4}{y(x)^{11}}(y(x)^6-x^6)\\
&=-70\frac{x^4y(x)}{(x^6-14)^2}
\end{align}
so yes, the first result is correct.

The second example works in the same fashion
\begin{align}
y(x)-\cos y(x)&=2x\\
y'(x)\,(1+\sin y(x))&=2\\
y''(x)\,(1+\sin y(x))+y'(x)^2\,\cos y(x)&=0\\
\implies\quad
y''(x)\,(1+\sin y(x))^3+4\,\cos y(x)&=0
\end{align}
