I just want to check am I practicing implicit differentiation correctly.
$$r = x+f(y+ rx + x^3)$$
where f is a differentiable function.
I am trying to use implicit differentiation to find $\frac{\partial r}{\partial x}$ in terms of $y,\ x,\ r$ and $f$.
$$\therefore\frac{\partial r}{\partial x} = \frac{\partial [x+f(y+ rx + x^3)]}{\partial x}$$
$$= 1+ \frac{\partial f(y+ rx + x^3)}{\partial x}$$
$$= 1+ f^\prime(y+ rx + x^3)\cdot\frac{\partial (y+ rx + x^3)}{\partial x}$$
$f^\prime$ replaces $f^\prime(y+ rx + x^3)$ for brevity.
$$=1 + f^\prime\cdot \left[ 0 + \frac{ \partial (rx) } { \partial x } + \frac{ \partial (x^3) } { \partial x } \right] $$
$$=1 + f^\prime\cdot \left[x\cdot\frac{ \partial r } { \partial x } + r + 3x^2 \right] $$
$$\therefore\frac{ \partial r } { \partial x } = 1 + f^\prime\cdot x\cdot\frac{ \partial r } { \partial x } + f^\prime\cdot r\ + f^\prime\cdot 3x^2 $$
$$\implies\frac{ \partial r } { \partial x }\cdot \left[1 - f^\prime\cdot x\ \right] = 1 + f^\prime\cdot\left[ r\ + 3x^2 \right] $$
$$\therefore\frac{ \partial r } { \partial x } = \frac{1 + f^\prime\cdot\left[ r\ + 3x^2 \right]}{\left[1 - f^\prime\cdot x\ \right]} $$
Is this the correct approach?