If $I$ is a maximal ideal in $\def\CC{\mathbb C}\CC[X]$, then there is an $\alpha\in\CC$ such that $I=(X-\alpha)$, and using this it is easy to see that $\dim_\CC I/I^2=1$.
On the other hand, the ideal $J=(X,Y)\subset A=\CC[X,Y]/(X^2-y^3)$ is maximal and $J/J^2$ is a vector space of dimension $2$.
It follows that $A$ is not isomorphic to $\CC[X]$.
Did you mean $x_i=g_i(x_1,x_2,\ldots,x_r)$? Even so, the answer is definitely no. Even if you have such substitutions, you cannot conclude that $
k[x_1, \dots, x_n] / \langle f_1, \dots, f_m \rangle \cong k[x_1, \dots, x_r]
$. For example, $k[x,y]/\langle x^2-y^2\rangle$ is not isomorphic as a ring to $k[x]$ even if you have a substitution $y=x$ that kills $x^2-y^2$.
However, if you have polynomials $f_i$ and $g_i$ for $i=1,2,\ldots,n-r$ such that $$f_i(x_1,x_2,\ldots,x_n)=x_{r+i}-g_i(x_1,x_2,\ldots,x_r)$$
or
$$f_i(x_1,x_2,\ldots,x_n)=x_{r+i}-g_i(x_1,x_2,\ldots,x_{r+i-1})\,,$$
then the answer is yes, namely,
$$k[x_1,x_2,\ldots,x_n]/\langle f_1,f_2,\ldots,f_{n-r}\rangle \cong k[x_1,x_2,\ldots,x_r]\,.$$ This is the case with your original problem.
Best Answer
Define
$$\phi: k[x,y]\to k[x],\;x\to x,\;\;y\to x^2,\text{ and expand accordingly}$$
or if you prefer: $\phi f(x,y):=f(x,x^2)$. Now check stuff.