Consider the function $f(t) = z(x(t),y(t))$, where
$$z(x,y) = x^5y^3, \quad x(t) = e^{t^2} \quad \text{and} \quad y(t) =e^{3t}.$$
Determine all the values of $t$ such that the tangent line to the graph of $f(t)$ is horizontal.
For no value, as $e^u>0$ for all $u$.
$t=-\dfrac{9}{5}$, $t=-\dfrac{9}{10}$ or $t=0.$
$t=\dfrac{-9-\sqrt{61}}{10}$, $t=-\dfrac{9}{10}$ or $\dfrac{-9+\sqrt{61}}{10}.$
$t=-\dfrac{9}{10}.$
Correct: Note that we have to find $t$ such that $f'(t)=0$. The partial derivatives of $z(x,y)$ at $(x(t),y(t)),$ the derivative of $x(t)$ and the derivative of $y(t)$ are:
$$
\begin{align*}
z'_x(x,y) &= y^3 \cdot 5x^4 = 5x^4y^3\\
z'_x(x(t),y(t)) &= 5\big(e^{t^2}\big)^4 \big(e^{3t}\big)^3 = 5e^{4t^2}e^{9t} = 5e^{4t^2 + 9t}\\[2mm]
z'_y(x,y) &= x^5 \cdot 3y^2 = 3x^5y^2\\
z'_y(x(t),y(t)) &= 3\big(e^{t^2}\big)^5 \big(e^{3t}\big)^2 = 3e^{5t^2}e^{6t} = 3e^{5t^2 + 6t}\\[2mm]
x'(t) &= e^{t^2} \cdot 2t = 2te^{t^2}\\[1mm]
y'(t) &= e^{3t} \cdot 3 = 3e^{3t}.
\end{align*}
$$
According to chain rule (case 1):
$$
f'(t) = 5e^{4t^2 + 9t} \cdot 2te^{t^2} + 3e^{5t^2 + 6t} \cdot 3e^{3t} = 10te^{5t^2+9t} + 9e^{5t^2 + 9t} = (10t+9)e^{5t^2+9t}.
$$
Finally, we have to solve $f'(t) =0$:
$$
\begin{array}{ccc}
{(10t+9)e^{5t^2+9t} = 0}\\
10t + 9 = 0 &\quad\mbox{ or }\quad& e^{5t^2 + 9t} = 0\\
10t =-9 && \text{not possible, because}\\
t = -\dfrac{9}{10} && e^u>0 \text{~for all~} u
\end{array}
$$
Go on.
Wrong: $x'(t) \neq e^{t^2}$ or you forgot to apply the chain rule: special case 1.
See Chain rule or Chain rule (case 1), Example 1 and Example 2.
Wrong: It is not true that $e^u=0$ if $u=0$.
Try again.
Wrong: It is not true that $e^u=0$ if $u=1$.
Try again.