To find $\frac{\partial f}{\partial t} \bigg|_{(t,w)=(1,1)}$ using the Chain Rule, let's follow these steps:

1. Express $f$ in terms of $t$ and $w$: $f(x, y) = e^{xy^3}$ where $x = t^2 + w^2 \quad \text{and} \quad y = tw^2.$

2. Compute the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$: $\frac{\partial f}{\partial x} = \frac{d}{dx}\left(e^{xy^3}\right) = y^3 e^{xy^3},$ $\frac{\partial f}{\partial y} = \frac{d}{dy}\left(e^{xy^3}\right) = 3xy^2 e^{xy^3}.$

3. Compute the partial derivatives $\frac{\partial x}{\partial t}$ and $\frac{\partial y}{\partial t}$: $\frac{\partial x}{\partial t} = \frac{d}{dt}(t^2 + w^2) = 2t,$ $\frac{\partial y}{\partial t} = \frac{d}{dt}(tw^2) = w^2.$

4. Apply the Chain Rule: $\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}.$

5. Substitute the values of the partial derivatives and the expressions for $f$, $x$, and $y$: $\frac{\partial f}{\partial t} = (y^3 e^{xy^3}) (2t) + (3xy^2 e^{xy^3}) (w^2).$

6. Evaluate at $t = 1$ and $w = 1$: $x = 1^2 + 1^2 = 2,$ $y = 1 \cdot 1^2 = 1.$ Therefore, $\frac{\partial f}{\partial x} \bigg|_{(t,w)=(1,1)} = 1^3 e^{2 \cdot 1^3} = e^2,$ $\frac{\partial f}{\partial y} \bigg|_{(t,w)=(1,1)} = 3 \cdot 2 \cdot 1^2 e^{2 \cdot 1^3} = 6e^2.$ Also, $\frac{\partial x}{\partial t} \bigg|_{(t,w)=(1,1)} = 2 \cdot 1 = 2,$ $\frac{\partial y}{\partial t} \bigg|_{(t,w)=(1,1)} = 1^2 = 1.$

7. Combine all these values: $\frac{\partial f}{\partial t} \bigg|_{(t,w)=(1,1)} = (e^2)(2) + (6e^2)(1) = 2e^2 + 6e^2 = 8e^2.$

So, $\frac{\partial f}{\partial t} \bigg|_{(t,w)=(1,1)} = 8e^2$.

Would you like more details or have any questions?

Here are 5 more questions you can ask:

1. How do you find the second derivative using the chain rule?
2. Can you explain the product rule with an example?
3. What is the integral of $e^{xy}$ with respect to $x$?
4. How do you solve differential equations using the chain rule?
5. What is the gradient of a multivariable function?

Tip: Always break down complex problems into smaller parts to simplify the solution process.