Find dy/dx at (1,1) for x^y + yx = sin(xy).

The given equation is:

$$x^y + yx = \sin(xy)$$

We need to find the derivative of each term.
For $x^y$, we use logarithmic differentiation or the fact that $x^y = e^{y \ln x}$:

$$\frac{d}{dx}(x^y) = \frac{d}{dx}(e^{y \ln x}) = e^{y \ln x} \cdot \frac{d}{dx}(y \ln x) = x^y \left( \frac{dy}{dx} \ln x + y \cdot \frac{1}{x} \right)$$

For $yx$, we use the product rule:

$$\frac{d}{dx}(yx) = \frac{dy}{dx} \cdot x + y \cdot 1 = x \frac{dy}{dx} + y$$

For $\sin(xy)$, we use the chain rule and product rule:

$$\frac{d}{dx}(\sin(xy)) = \cos(xy) \cdot \frac{d}{dx}(xy) = \cos(xy) \left( \frac{dy}{dx} \cdot x + y \cdot 1 \right) = \cos(xy) \left( x \frac{dy}{dx} + y \right)$$

Recall that $\ln(1) = 0$.

$$1^1 \left( \frac{dy}{dx} \ln 1 + \frac{1}{1} \right) + 1 \cdot \frac{dy}{dx} + 1 = \cos(1 \cdot 1) \left( 1 \cdot \frac{dy}{dx} + 1 \right)$$ $$1 \left( \frac{dy}{dx} \cdot 0 + 1 \right) + \frac{dy}{dx} + 1 = \cos(1) \left( \frac{dy}{dx} + 1 \right)$$

$$1 + \frac{dy}{dx} + 1 = \cos(1) \frac{dy}{dx} + \cos(1)$$ $$2 + \frac{dy}{dx} = \cos(1) \frac{dy}{dx} + \cos(1)$$

Subtract $\cos(1) \frac{dy}{dx}$ from both sides and subtract 2 from both sides:

$$\frac{dy}{dx} - \cos(1) \frac{dy}{dx} = \cos(1) - 2$$

Factor out $\frac{dy}{dx}$:

$$\frac{dy}{dx} (1 - \cos(1)) = \cos(1) - 2$$

Divide by $(1 - \cos(1))$:

$$\frac{dy}{dx} = \frac{\cos(1) - 2}{1 - \cos(1)}$$

The value of $\frac{dy}{dx}$ at the point $(1,1)$ is:

$$\frac{\cos(1) - 2}{1 - \cos(1)}$$