Evaluating $y'$ for $y - x^{2}y^{2} - \cos(xy) = 4$ I want to evaluate the title's implicit differentiation. Here are the steps I've taken:
1. Derive both sides of the equation
$$\frac {d}{dx}(y - x^2y^2 - cos(xy)) = \frac {d}{dx}(4)$$
2. Note that the derivative of a constant is 0
$$\frac {d}{dx}(y - x^2y^2 - cos(xy)) = 0$$
3. Apply rule of difference
$$\frac {d}{dx}(y) - \frac {d}{dx}(x^2y^2) - \frac {d}{dx}(cos(xy)) = 0$$
4. Apply chain rule
$$\frac {d}{dx}(y) - \frac {d}{dx}(x^2y^2) - \left(\frac {d}{dx}cos(xy)*\frac {d}{dx}(xy)\right) = 0$$
5. Apply product rule
$$\frac {d}{dx}(y) - \left(\frac {d}{dx}(x^2)y^2 + x^2\frac {d}{dx}(y^2)\right) - \left(\frac {d}{dx}cos(xy)*\left(\frac {d}{dx}(x)y + x\frac {d}{dx} (y)\right)\right) = 0$$
6. Solve derivatives
$$\frac {d}{dx}(y) - \left(2xy^2 + 2x^2y\frac {dy}{dx}\right) - \left(-sin(xy)*\left(1y + x\frac {d}{dx} (y)\right)\right) = 0$$
7. Factorize and isolate the derivative of y wrt x
$$\frac {dy}{dx}(1-2x^2y+xsin(xy)) = 2xy^2 - ysin(xy) \Rightarrow$$
$$\frac {dy}{dx} = \frac {2xy^2 - ysin(xy)}{1+2x^2y+xsin(xy)}$$
Is this correct? And if so, is this the best approach? Also, how can I improve my mathematical dissertation?

Hey I see your solution is almost correct, but you got the sign wrong. Here is the similar method. 

*

*First, we derive the following equation via implicit differentiation.
\begin{equation*} \frac{d}{dx} \left( y-x^{2}y^{2} -\cos(xy) = 4 \right) = \frac{d}{dx}(4) \end{equation*}

*Second, we sum up the term differentiate and manager the constant factor.
\begin{equation*} \left( - \left( \frac{d}{dx} \cos(xy) \right) + \frac{d}{dx}(y)-\frac{d}{dx}(x^{2}y^{2}) \right) =\frac{d}{dx}(4) \end{equation*}

*We apply the chain rule, $\frac{d}{dx}(\cos(xy))=\frac{d\cos(u)}{du}\frac{du}{dx}$, where $u=xy$ and $\frac{d}{dx}(\cos(u))=-sin(u) $
, we get
\begin{align*} \frac{d}{dx}(y)-\frac{d}{dx}(x^{2}y^{2})+\frac{d}{dx}(xy)\sin(xy) &= \frac{d}{dx}(4)  \\
-\left(\frac{d}{dx}(x^{2}y^{2} \right)+\sin(xy)\left(x\left(\frac{d}{dx}(y) \right) + \left( \frac{d}{dx}(x) \right) y \right)+y'(x) &= \frac{d}{dx}(4) \\
\end{align*}

*Use product rule, $\frac{d}{dx}(uv) = v\frac{du}{dx}+u\frac{dv}{dx}$, where $u=x^{2}$ and $v=y^{2}$:
\begin{align*} -\left( x^{2} \frac{d}{dx}(y^{2}) + y^{2}\frac{d}{dx}(x^{2})\right) + \sin(xy)\left(x\frac{d}{dx}(y)+y\frac{d}{dx}(x) \right) y'(x) &= \frac{d}{dx}(4) \end{align*}.

*After the whole substitution, we arrived at
\begin{equation*} (1+x\sin(xy)-2x^{2}y)y'(x) =-sin(xy)y+2xy^{2} \end{equation*}

*Divide both side by $-2x^{2}y+xsin(xy)+1 $
\begin{equation*} y'(x) = \frac{-sin(xy)y+2xy^{2}}{1+xsin(xy)-2x^{2}y} \end{equation*}
Well, just carefully with the sign!
