Replacing in equation introduces more solutions

Suppose first that $f$ is a real-valued function of one variable. The equation $$ x = f(x) \tag{1} $$ acts as a condition, selecting values of $x$ for which (1) is true. Substituting (1) into itself gives a new condition, $$ x = f\bigl(f(x)\bigr) = f^{[2]}(x), \tag{2} $$ and so forth.

Certainly every solution of (1) is a solution of (2). If the function $f$ is not injective (one-to-one), however, (2) can have solutions that are not solutions of (1).

For example, if $f(x) = 4x(1 - x)$, then $f$ maps $[0, 1]$ onto $[0, 1]$. Each point except $x = \frac{1}{2}$ has two preimages, and $f$ maps each interval $[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$ bijectively to $[0, 1]$. It follows that $f \circ f$ maps each half-interval onto $[0, 1]$, and each point of $[0, 1]$ has two preimages in each half-interval, so $f^{[2]} = f \circ f$ has four fixed points, etc. (Diagram below.) In this example, the $n$-fold composition of $f$ with itself, $f^{[n]}$, has $2^{n}$ fixed points, i.e., the equation $$ x = f^{[n]}(x) = (\underbrace{f \circ \dots \circ f}_{n \text{times}})(x) \tag{n} $$ has $2^{n}$ solutions, even though (n) is obtained from (1) by successively substituting (1) into itself.

Your situation is analogous: Starting from $$ y = 2 - x^{2} - y^{2} = f(x, y), \tag{1a} $$ you substitute (1a) into itself, obtaining $$ y = 2 - x^{2} - f(x, y)^{2} = f\bigl(x, f(x, y)\bigr), \tag{2a} $$ and so forth.

In your example, $f$ has qualitatively similar behavior along the $y$-axis to the "logistic map" $f(x) = 4x(1 - x)$, and the solution sets $$ y = f\Bigl(\dots f\bigl(x, f(x, y)\bigr)\dots\Bigr) $$ become increasingly complicated with successive iteration.

This type of phenomenon (e.g., the precise location/shape of the solutions) is generally complicated (chaotic in the technical sense). Wikipedia pages of possible interest include:

• The logistic map
• The tent map
• The horseshoe map
• Dynamical systems