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The problem asks to compute the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\), given that \( z \) is defined implicitly by the equation

\[
2g(z) + f(x^2, y^2) = 0,
\]

where \( f \) is a function with continuous second-order derivatives, and \( g \) is a twice-differentiable function.

### Step-by-step approach:
We are dealing with implicit differentiation, where \( z = z(x, y) \). Therefore, \( z \) depends on both \( x \) and \( y \), and we will apply the chain rule.

Let's begin by implicitly differentiating the given equation with respect to \( x \) and then with respect to \( y \).

#### 1. Implicit differentiation with respect to \( x \):
\[
\frac{d}{dx} \left( 2g(z) + f(x^2, y^2) \right) = 0
\]
First, apply the chain rule to each term:

- For \( 2g(z) \), we have:
  \[
  2g'(z) \frac{\partial z}{\partial x}
  \]
- For \( f(x^2, y^2) \), differentiate with respect to \( x \):
  \[
  \frac{\partial}{\partial x} f(x^2, y^2) = \frac{\partial f}{\partial x^2} \cdot 2x
  \]
Thus, the full derivative with respect to \( x \) becomes:
\[
2g'(z) \frac{\partial z}{\partial x} + 2x \frac{\partial f}{\partial (x^2)} = 0
\]
Solve for \( \frac{\partial z}{\partial x} \):
\[
\frac{\partial z}{\partial x} = -\frac{x \frac{\partial f}{\partial (x^2)}}{g'(z)}
\]

#### 2. Implicit differentiation with respect to \( y \):
Now differentiate the equation with respect to \( y \):
\[
\frac{d}{dy} \left( 2g(z) + f(x^2, y^2) \right) = 0
\]
Apply the chain rule again:

- For \( 2g(z) \), we get:
  \[
  2g'(z) \frac{\partial z}{\partial y}
  \]
- For \( f(x^2, y^2) \), differentiate with respect to \( y \):
  \[
  \frac{\partial}{\partial y} f(x^2, y^2) = \frac{\partial f}{\partial y^2} \cdot 2y
  \]
Thus, the full derivative with respect to \( y \) becomes:
\[
2g'(z) \frac{\partial z}{\partial y} + 2y \frac{\partial f}{\partial (y^2)} = 0
\]
Solve for \( \frac{\partial z}{\partial y} \):
\[
\frac{\partial z}{\partial y} = -\frac{y \frac{\partial f}{\partial (y^2)}}{g'(z)}
\]

### Final answers:
\[
\frac{\partial z}{\partial x} = -\frac{x \frac{\partial f}{\partial (x^2)}}{g'(z)}
\]
\[
\frac{\partial z}{\partial y} = -\frac{y \frac{\partial f}{\partial (y^2)}}{g'(z)}
\]

Would you like to explore these results further or ask for clarification? Here are some related questions to expand on this topic:

1. How would the result change if \( f(x^2, y^2) \) were a more complex function, such as a sum of different powers?
2. What happens if \( g'(z) = 0 \)? How would this affect the derivatives?
3. Can we generalize this procedure for higher-dimensional implicit functions?
4. How does the chain rule for implicit differentiation differ when dealing with multivariable functions like this?
5. Could we apply this method to solve for higher-order partial derivatives, such as \(\frac{\partial^2 z}{\partial x^2}\)?

**Tip:** When dealing with implicit differentiation, always carefully apply the chain rule, ensuring that each term dependent on multiple variables is treated accordingly.

Sea f una función con derivadas de segundo orden continuas y g una función dos veces derivable. Supongamos que la ecuación 2g(z) + f(x^2, y^2) = 0 define a z como función implícita de x y y. Calcule ∂z/∂x y ∂z/∂y.

Learn how to calculate the partial derivatives ∂z/∂x and ∂z/∂y from the implicit equation 2g(z) + f(x^2, y^2) = 0 using the chain rule. This multivariable calculus problem involves implicit differentiation and the use of continuous second-order derivatives.

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