How to solve the following system of simultaneous equations for $f(y)$ and $g(y)$? $$ \begin{align} y^4 &= f(y) + y^2 g \left( y^2 \right), \tag{1} \newline y^2 &= -2y f^\prime (y) + f(y) - y^4 g^\prime \left( y^2 \right). \tag{2} \end{align} $$
My Attempt:
Upon differentiating both the sides of the first equation w.r.t $y$, we obtain $$ \begin{align} y^4 &= f(y) + y^2 g \left( y^2 \right), \tag{1} \newline y^2 &= -2y f^\prime (y) + f(y) - y^4 g^\prime \left( y^2 \right), \tag{2} \newline 4y^3 &= f^\prime(y) + 2y g\left( y^2 \right) + 2y^3 g^\prime \left( y^2 \right). \tag{3} \end{align} $$
Now multiplying (3) by $\frac{y}{2}$ we get $$ \begin{align} y^4 &= f(y) + y^2 g \left( y^2 \right), \tag{1} \newline y^2 &= -2y f^\prime (y) + f(y) - y^4 g^\prime \left( y^2 \right), \tag{2} \newline 4y^3 &= f^\prime(y) + 2y g\left( y^2 \right) + 2y^3 g^\prime \left( y^2 \right), \tag{3} \newline 2y^4 &= \frac{y}{2}f^\prime(y) + y^2 g\left( y^2 \right) + y^4 g^\prime \left( y^2 \right). \tag{4} \end{align} $$
Now adding (2) to (4) we get $$ \begin{align} y^4 &= f(y) + y^2 g \left( y^2 \right), \tag{1} \newline y^2 &= -2y f^\prime (y) + f(y) - y^4 g^\prime \left( y^2 \right), \tag{2} \newline 4y^3 &= f^\prime(y) + 2y g\left( y^2 \right) + 2y^3 g^\prime \left( y^2 \right), \tag{3} \newline 2y^4 &= \frac{y}{2}f^\prime(y) + y^2 g\left( y^2 \right) + y^4 g^\prime \left( y^2 \right), \tag{4} \newline 2y^4 + y^2 &= - \frac{3y}{2} f^\prime(y) + f(y) + y^2 g\left( y^2 \right). \tag{5} \end{align} $$
Now using (1) in (5) we obtain $$ \begin{align} y^4 &= f(y) + y^2 g \left( y^2 \right), \tag{1} \newline y^2 &= -2y f^\prime (y) + f(y) - y^4 g^\prime \left( y^2 \right), \tag{2} \newline 4y^3 &= f^\prime(y) + 2y g\left( y^2 \right) + 2y^3 g^\prime \left( y^2 \right), \tag{3} \newline 2y^4 &= \frac{y}{2}f^\prime(y) + y^2 g\left( y^2 \right) + y^4 g^\prime \left( y^2 \right), \tag{4} \newline 2y^4 + y^2 &= - \frac{3y}{2} f^\prime(y) + f(y) + y^2 g\left( y^2 \right), \tag{5} \newline 2y^4 + y^2 &= - \frac{3y}{2} f^\prime(y) + y^4. \tag{6} \end{align} $$
And, upon simplifying (6) we obtain $$ \begin{align} y^4 &= f(y) + y^2 g \left( y^2 \right), \tag{1} \newline y^2 &= -2y f^\prime (y) + f(y) - y^4 g^\prime \left( y^2 \right), \tag{2} \newline 4y^3 &= f^\prime(y) + 2y g\left( y^2 \right) + 2y^3 g^\prime \left( y^2 \right), \tag{3} \newline 2y^4 &= \frac{y}{2}f^\prime(y) + y^2 g\left( y^2 \right) + y^4 g^\prime \left( y^2 \right), \tag{4} \newline 2y^4 + y^2 &= - \frac{3y}{2} f^\prime(y) + f(y) + y^2 g\left( y^2 \right), \tag{5} \newline y^4 + y^2 &= - \frac{3y}{2} f^\prime(y). \tag{6} \end{align} $$
Solving (6) for $f^\prime(y)$, we obtain $$ \begin{align} y^4 &= f(y) + y^2 g \left( y^2 \right), \tag{1} \newline y^2 &= -2y f^\prime (y) + f(y) - y^4 g^\prime \left( y^2 \right), \tag{2} \newline 4y^3 &= f^\prime(y) + 2y g\left( y^2 \right) + 2y^3 g^\prime \left( y^2 \right), \tag{3} \newline 2y^4 &= \frac{y}{2}f^\prime(y) + y^2 g\left( y^2 \right) + y^4 g^\prime \left( y^2 \right), \tag{4} \newline 2y^4 + y^2 &= - \frac{3y}{2} f^\prime(y) + f(y) + y^2 g\left( y^2 \right), \tag{5} \newline f^\prime(y) &= -\frac{2}{3} \left( y^3 + y \right). \tag{6} \end{align} $$
Upon integrating (6) w.r.t to $y$, we get $$ \begin{align} y^4 &= f(y) + y^2 g \left( y^2 \right), \tag{1} \newline y^2 &= -2y f^\prime (y) + f(y) - y^4 g^\prime \left( y^2 \right), \tag{2} \newline 4y^3 &= f^\prime(y) + 2y g\left( y^2 \right) + 2y^3 g^\prime \left( y^2 \right), \tag{3} \newline 2y^4 &= \frac{y}{2}f^\prime(y) + y^2 g\left( y^2 \right) + y^4 g^\prime \left( y^2 \right), \tag{4} \newline 2y^4 + y^2 &= - \frac{3y}{2} f^\prime(y) + f(y) + y^2 g\left( y^2 \right), \tag{5} \newline f^\prime(y) &= -\frac{2}{3} \left( y^3 + y \right), \tag{6} \newline f(y) &= - \frac{y^4}{6} - \frac{y^2}{3} + c, \tag{7} \end{align} $$ where $c$ is a constant (independent of $y$).
Substituting the value of $f(y)$ from (7) into (1) we obtain $$ \begin{align} y^4 &= - \frac{y^4}{6} - \frac{y^2}{3} + c + y^2 g \left( y^2 \right), \tag{1} \newline y^2 &= -2y f^\prime (y) + f(y) - y^4 g^\prime \left( y^2 \right), \tag{2} \newline 4y^3 &= f^\prime(y) + 2y g\left( y^2 \right) + 2y^3 g^\prime \left( y^2 \right), \tag{3} \newline 2y^4 &= \frac{y}{2}f^\prime(y) + y^2 g\left( y^2 \right) + y^4 g^\prime \left( y^2 \right), \tag{4} \newline 2y^4 + y^2 &= - \frac{3y}{2} f^\prime(y) + f(y) + y^2 g\left( y^2 \right), \tag{5} \newline f^\prime(y) &= -\frac{2}{3} \left( y^3 + y \right), \tag{6} \newline f(y) &= - \frac{y^4}{6} - \frac{y^2}{3} + c, \tag{7} \end{align} $$
Solving (1) for $g\left( y^2 \right)$, we get $$ \begin{align} g \left( y^2 \right) &= \frac{7y^2}{6} + \frac{1}{3} - \frac{c}{y^2} , \tag{1} \newline y^2 &= -2y f^\prime (y) + f(y) - y^4 g^\prime \left( y^2 \right), \tag{2} \newline 4y^3 &= f^\prime(y) + 2y g\left( y^2 \right) + 2y^3 g^\prime \left( y^2 \right), \tag{3} \newline 2y^4 &= \frac{y}{2}f^\prime(y) + y^2 g\left( y^2 \right) + y^4 g^\prime \left( y^2 \right), \tag{4} \newline 2y^4 + y^2 &= - \frac{3y}{2} f^\prime(y) + f(y) + y^2 g\left( y^2 \right), \tag{5} \newline f^\prime(y) &= -\frac{2}{3} \left( y^3 + y \right), \tag{6} \newline f(y) &= - \frac{y^4}{6} - \frac{y^2}{3} + c. \tag{7} \end{align} $$
Finally, from (7) and (1) we get $$ \begin{align} f(y) &= - \frac{y^4}{6} - \frac{y^2}{3} + c, \newline g\left(y^2 \right) &= \frac{7y^2}{6} + \frac{1}{3} - \frac{c}{y^2}, \newline & \mbox{ and upon substituting $y$ for $y^2$ in the formula for $g(y^2)$ we get } \newline g(y) &= \frac{7y}{6} + \frac{1}{3} - \frac{c}{y}. \end{align} $$
Therefore our required solution is $$ f(y) = - \frac{y^4}{6} - \frac{y^2}{3} + c, \qquad \mbox{ and } \qquad g(y) = \frac{7y}{6} + \frac{1}{3} - \frac{c}{y}. $$
Is my solution correct and accurate in each and every detail? If so, have I managed to end up with a correct pair of formulas for $f(y)$ and $g(y)$? If not, then where are the errors or issues?