Find all functions $f:(1,+\infty) \rightarrow (1,+\infty)$ that satisfy the following condition:
for arbitrary $x,y>1$ and $u,v>0$, inequality $f(x^uy^v)\le f(x)^{\dfrac{1}{4u}}f(y)^{\dfrac{1}{4v}}$ holds.
1. **Define the function \( g \):**
   Consider a function \( g \) such that \( f(x) = g(x)^{\frac{1}{\ln x}} \). This transformation is valid for any \( x > 1 \) and \( g(x) \) is unique.

2. **Substitute \( y = x \) in the given inequality:**
   Given the inequality \( f(x^u y^v) \le f(x)^{\frac{1}{4u}} f(y)^{\frac{1}{4v}} \), substitute \( y = x \):
   \[
   f(x^{u+v}) \le f(x)^{\frac{1}{4u}} f(x)^{\frac{1}{4v}}
   \]
   Using the definition of \( f \), we get:
   \[
   g(x^{u+v})^{\frac{1}{(u+v) \ln x}} \le \left( g(x)^{\frac{1}{\ln x}} \right)^{\frac{1}{4u}} \left( g(x)^{\frac{1}{\ln x}} \right)^{\frac{1}{4v}}
   \]

3. **Simplify the inequality:**
   Simplify the right-hand side:
   \[
   g(x^{u+v})^{\frac{1}{(u+v) \ln x}} \le g(x)^{\frac{1}{4u \ln x}} g(x)^{\frac{1}{4v \ln x}}
   \]
   Combine the exponents on the right-hand side:
   \[
   g(x^{u+v})^{\frac{1}{(u+v) \ln x}} \le g(x)^{\frac{1}{4u \ln x} + \frac{1}{4v \ln x}}
   \]
   Since \( \frac{1}{4u \ln x} + \frac{1}{4v \ln x} = \frac{1}{4 \ln x} \left( \frac{1}{u} + \frac{1}{v} \right) \), we have:
   \[
   g(x^{u+v})^{\frac{1}{(u+v) \ln x}} \le g(x)^{\frac{1}{4 \ln x} \left( \frac{1}{u} + \frac{1}{v} \right)}
   \]

4. **Exponentiate both sides:**
   Raise both sides to the power of \( (u+v) \ln x \):
   \[
   g(x^{u+v}) \le g(x)^{\frac{(u+v)}{4} \left( \frac{1}{u} + \frac{1}{v} \right)}
   \]
   Simplify the exponent on the right-hand side:
   \[
   g(x^{u+v}) \le g(x)^{\frac{(u+v)}{4} \left( \frac{1}{u} + \frac{1}{v} \right)} = g(x)^{\frac{(u+v)}{4} \left( \frac{u+v}{uv} \right)} = g(x)^{\frac{(u+v)^2}{4uv}}
   \]

5. **Special case \( u = v \):**
   Let \( u = v \):
   \[
   g(x^{2u}) \le g(x)^{\frac{(2u)^2}{4u^2}} = g(x)
   \]
   This implies:
   \[
   g(x^{2u}) \le g(x)
   \]
   Since this holds for any \( u > 0 \), \( g \) must be a constant function. Let \( g(x) = c \) for some constant \( c > 1 \).

6. **Substitute back to find \( f \):**
   Since \( g(x) = c \), we have:
   \[
   f(x) = c^{\frac{1}{\ln x}}
   \]

7. **Verify the solution:**
   Check that \( f(x) = c^{\frac{1}{\ln x}} \) satisfies the original inequality:
   \[
   f(x^u y^v) = c^{\frac{1}{\ln(x^u y^v)}} = c^{\frac{1}{u \ln x + v \ln y}}
   \]
   And:
   \[
   f(x)^{\frac{1}{4u}} f(y)^{\frac{1}{4v}} = \left( c^{\frac{1}{\ln x}} \right)^{\frac{1}{4u}} \left( c^{\frac{1}{\ln y}} \right)^{\frac{1}{4v}} = c^{\frac{1}{4u \ln x}} c^{\frac{1}{4v \ln y}} = c^{\frac{1}{4u \ln x} + \frac{1}{4v \ln y}}
   \]
   Since \( \frac{1}{u \ln x + v \ln y} \le \frac{1}{4u \ln x} + \frac{1}{4v \ln y} \), the inequality holds.

The final answer is \( \boxed{ f(x) = c^{\frac{1}{\ln x}} } \) where \( c > 1 \).