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To find the equation of the tangent plane to the surface defined by:

\[
xy^2 + yz^2 - zx^2 = 1
\]

at the point \((1, 1, 1)\), we will follow these steps:

### Step 1: Compute the Gradient of the Implicit Function
The surface can be viewed as an implicit function \( F(x, y, z) = xy^2 + yz^2 - zx^2 - 1 \). The gradient of \(F\) is the vector of partial derivatives with respect to \(x\), \(y\), and \(z\):

\[
\nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)
\]

1. **Partial derivative with respect to \(x\)**:
   \[
   \frac{\partial F}{\partial x} = y^2 - z^2x
   \]

2. **Partial derivative with respect to \(y\)**:
   \[
   \frac{\partial F}{\partial y} = 2xy + z^2
   \]

3. **Partial derivative with respect to \(z\)**:
   \[
   \frac{\partial F}{\partial z} = 2yz - 2xz
   \]

### Step 2: Evaluate the Gradient at the Point \((1, 1, 1)\)
Substitute \(x = 1\), \(y = 1\), and \(z = 1\) into the partial derivatives:

1. For \( \frac{\partial F}{\partial x} \):
   \[
   \frac{\partial F}{\partial x} = 1^2 - 1^2 \times 1 = 1 - 1 = 0
   \]

2. For \( \frac{\partial F}{\partial y} \):
   \[
   \frac{\partial F}{\partial y} = 2 \times 1 \times 1 + 1^2 = 2 + 1 = 3
   \]

3. For \( \frac{\partial F}{\partial z} \):
   \[
   \frac{\partial F}{\partial z} = 2 \times 1 \times 1 - 2 \times 1 \times 1 = 2 - 2 = 0
   \]

Thus, the gradient at \((1, 1, 1)\) is:

\[
\nabla F(1, 1, 1) = (0, 3, 0)
\]

### Step 3: Equation of the Tangent Plane
The equation of the tangent plane at \((1, 1, 1)\) is given by:

\[
\nabla F(1, 1, 1) \cdot \left( (x - 1), (y - 1), (z - 1) \right) = 0
\]

Substituting the gradient \((0, 3, 0)\), this becomes:

\[
0(x - 1) + 3(y - 1) + 0(z - 1) = 0
\]

Simplifying:

\[
3(y - 1) = 0
\]

\[
y - 1 = 0
\]

Thus, the equation of the tangent plane is simply:

\[
y = 1
\]

However, based on the image's structure, the expected plane form seems to be modified. Could you confirm this or clarify the image further? Feel free to ask!

Find the equation of the tangent plane to the surface xy^2 + yz^2 - zx^2 = 1 at the point (1, 1, 1).

This problem explores how to find the tangent plane to the surface xy^2 + yz^2 - zx^2 = 1 at the point (1, 1, 1). The solution involves calculating partial derivatives and using the gradient formula for tangent planes in multivariable calculus.

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