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Let's solve the problem step by step.

### Given:
The surface is defined by:
\[
xy^2 + yz^2 - zx^2 = 1
\]
We need to find the equation of the tangent plane at the point \( (1, 1, 1) \).

### Step 1: Compute the gradient of the surface
To find the equation of the tangent plane, we first need the gradient vector of the surface. The gradient of the surface \( F(x, y, z) = xy^2 + yz^2 - zx^2 - 1 \) is given by:
\[
\nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)
\]

#### Partial derivatives:
1. \( \frac{\partial F}{\partial x} = y^2 - z^2 \)
2. \( \frac{\partial F}{\partial y} = 2xy + z^2 \)
3. \( \frac{\partial F}{\partial z} = 2yz - 2zx \)

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

1. \( \frac{\partial F}{\partial x} \bigg|_{(1,1,1)} = 1^2 - 1^2 = 0 \)
2. \( \frac{\partial F}{\partial y} \bigg|_{(1,1,1)} = 2(1)(1) + 1^2 = 2 + 1 = 3 \)
3. \( \frac{\partial F}{\partial z} \bigg|_{(1,1,1)} = 2(1)(1) - 2(1)(1) = 2 - 2 = 0 \)

So, the gradient at the point \( (1, 1, 1) \) is:
\[
\nabla F(1, 1, 1) = (0, 3, 0)
\]

### Step 3: Equation of the tangent plane
The equation of the tangent plane at a point \( (x_0, y_0, z_0) \) on the surface is given by:
\[
\nabla F(x_0, y_0, z_0) \cdot (x - x_0, y - y_0, z - z_0) = 0
\]
At \( (1, 1, 1) \), the equation becomes:
\[
(0, 3, 0) \cdot (x - 1, y - 1, z - 1) = 0
\]

This simplifies to:
\[
0(x - 1) + 3(y - 1) + 0(z - 1) = 0
\]
\[
3(y - 1) = 0
\]
\[
y = 1
\]

So, the equation of the tangent plane is:
\[
y = 1
\]

The values for the tangent plane expression are \( x + 3(y - 1) + z = 0 \). Would you like more details on any step?

### 5 Related Questions:
1. How do you compute the gradient of a multivariable function?
2. What is the general formula for the equation of a tangent plane?
3. Why is the gradient vector perpendicular to the tangent plane?
4. How would the equation change if the point was different, say \( (2, 1, 1) \)?
5. How can you visualize the tangent plane in 3D space?

### Tip:
The gradient of a function always points in the direction of the steepest ascent, and its magnitude tells you how steep the slope is.

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

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 involves finding partial derivatives and using the gradient to compute the tangent plane equation, suitable for undergraduate calculus students.

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