Given that:

• $u = f(t)$, where $t = ax + by + cz$,
  and you are asked to prove the following relationships involving partial derivatives: $bc \frac{\partial u}{\partial x} = ac \frac{\partial u}{\partial y} = ab \frac{\partial u}{\partial z}.$

Approach:

1. Chain Rule Application:

   We are tasked with differentiating $u$ with respect to $x$, $y$, and $z$, while $t = ax + by + cz$ depends on these variables. Since $u = f(t)$, we use the chain rule for multivariable functions. The chain rule tells us that: $\frac{\partial u}{\partial x} = \frac{d f(t)}{d t} \cdot \frac{\partial t}{\partial x}.$ Similarly, we have: $\frac{\partial u}{\partial y} = \frac{d f(t)}{d t} \cdot \frac{\partial t}{\partial y}, \quad \frac{\partial u}{\partial z} = \frac{d f(t)}{d t} \cdot \frac{\partial t}{\partial z}.$

2. Computing Partial Derivatives of $t$:

   Recall that $t = ax + by + cz$. Therefore:

   • $\frac{\partial t}{\partial x} = a$,
   • $\frac{\partial t}{\partial y} = b$,
   • $\frac{\partial t}{\partial z} = c$.

3. Substituting into Chain Rule:

   Using the derivatives of $t$ in the chain rule expressions, we obtain: $\frac{\partial u}{\partial x} = \frac{d f(t)}{d t} \cdot a,$ $\frac{\partial u}{\partial y} = \frac{d f(t)}{d t} \cdot b,$ $\frac{\partial u}{\partial z} = \frac{d f(t)}{d t} \cdot c.$

4. Relating the Equations:

   Now we can express the relationships: $bc \frac{\partial u}{\partial x} = bc \cdot \frac{d f(t)}{d t} \cdot a = abc \cdot \frac{d f(t)}{d t},$ $ac \frac{\partial u}{\partial y} = ac \cdot \frac{d f(t)}{d t} \cdot b = abc \cdot \frac{d f(t)}{d t},$ $ab \frac{\partial u}{\partial z} = ab \cdot \frac{d f(t)}{d t} \cdot c = abc \cdot \frac{d f(t)}{d t}.$

   Therefore, we can conclude: $bc \frac{\partial u}{\partial x} = ac \frac{\partial u}{\partial y} = ab \frac{\partial u}{\partial z} = abc \cdot \frac{d f(t)}{d t}.$

Conclusion:

We have proven the required relationships, showing that: $bc \frac{\partial u}{\partial x} = ac \frac{\partial u}{\partial y} = ab \frac{\partial u}{\partial z}.$

Would you like further clarification on any step? Here are five related questions for deeper understanding:

1. How does the chain rule work when dealing with multivariable functions?
2. Can we extend this proof to higher-order partial derivatives of $u$?
3. What would happen if the coefficients $a$, $b$, and $c$ were functions of $x$, $y$, and $z$?
4. How does this result connect to the concept of directional derivatives in multivariable calculus?
5. Can you prove similar identities using other types of coordinate transformations?

Tip: When applying the chain rule, always carefully track the dependency of each variable in the function. This will ensure you apply the correct partial derivatives and achieve the desired results.